Selected Solutions to Munkres’s Topology, 2nd Ed.

Takumi Murayama

December 20, 2014

These solutions are the result of taking MAT365 Topology in the Fall of 2012 atPrinceton University. This is not a complete set of solutions; see the List of SolvedExercises at the end. Please e-mail takumim@umich.edu with any corrections.

Contents

I General Topology 3

1 Set Theory and Logic 37 Countable and Uncountable Sets . . . . . . . . . . . . . . . . . . . . 3

2 Topological Spaces and Continuous Functions 313 Basis for a Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 The Subspace Topology . . . . . . . . . . . . . . . . . . . . . . . . . 517 Closed Sets and Limit Points . . . . . . . . . . . . . . . . . . . . . . 618 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 819 The Product Topology . . . . . . . . . . . . . . . . . . . . . . . . . 920 The Metric Topology . . . . . . . . . . . . . . . . . . . . . . . . . . 1021 The Metric Topology (continued) . . . . . . . . . . . . . . . . . . . . 1522 The Quotient Topology . . . . . . . . . . . . . . . . . . . . . . . . . 18

3 Connectedness and Compactness 2023 Connected Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2024 Connected Subspaces of the Real Line . . . . . . . . . . . . . . . . . 2125 Components and Local Connectedness . . . . . . . . . . . . . . . . . 2627 Compact Subspaces of the Real Line . . . . . . . . . . . . . . . . . . 2729 Local Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1

4 Countability and Separation Axioms 2830 The Countability Axioms . . . . . . . . . . . . . . . . . . . . . . . . 2831 The Separation Axioms . . . . . . . . . . . . . . . . . . . . . . . . . 3032 Normal Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3133 The Urysohn Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . 3234 The Urysohn Metrization Theorem . . . . . . . . . . . . . . . . . . . 3236 Imbeddings of Manifolds . . . . . . . . . . . . . . . . . . . . . . . . 33

II Algebraic Topology 35

9 The Fundamental Group 3551 Homotopy of Paths . . . . . . . . . . . . . . . . . . . . . . . . . . . 3552 The Fundamental Group . . . . . . . . . . . . . . . . . . . . . . . . 3653 Covering Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3754 The Fundamental Group of the Circle . . . . . . . . . . . . . . . . . 3858 Deformation Retracts and Homotopy Type . . . . . . . . . . . . . . 3959 The Fundamental Group of Sn . . . . . . . . . . . . . . . . . . . . . 4360 Fundamental Groups of Some Surfaces . . . . . . . . . . . . . . . . . 44

11 The Seifert-van Kampen Theorem 4567 Direct Sums of Abelian Groups . . . . . . . . . . . . . . . . . . . . . 4568 Free Products of Groups . . . . . . . . . . . . . . . . . . . . . . . . 4571 The Fundamental Group of a Wedge of Circles . . . . . . . . . . . . 4673 The Fundamental Groups of the Torus and the Dunce Cap . . . . . 48

12 Classification of Surfaces 4974 Fundamental Groups of Surfaces . . . . . . . . . . . . . . . . . . . . 4975 Homology of Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2

Part I

General Topology

1 Set Theory and Logic

7 Countable and Uncountable Sets

Exercise 7.5. Determine, for each of the following sets, whether or not it is count-able. Justify your answers.

(j) The set J of all finite subsets of Z+.

Solution for (j). We claim J is countable. Consider I =⋃∞n=0 In where In is the set

of sequences with n elements. Each In is countable by Theorem 7.6 so I is countableby Theorem 7.5. Identifying each finite subset in J with the finite sequence withthe same elements in increasing order, we see that J ⊆ I, and so J is countable byCorollary 7.3.

2 Topological Spaces and Continuous Functions

13 Basis for a Topology

Exercise 13.3. Show that the collection Tc given in Example 4 of §12 is a topologyon the set X. Is the collection

T∞ = U | X \ U is infinite or empty or all of X

a topology on X?

Proof. Recall Example 12.4: Let X be a set; let Tc be the collection of all subsetsU of X such that X \ U either is countable or is all of X. We claim this forms atopology on X; we will follow the numbering for the definition of a topology on p. 76.(1) X \ ∅ = X and X \ X = ∅ is countable; (2) X \

⋃α∈A Uα =

⋂α∈A(X \ Uα) is

countable since it is an intersection of countable sets, unless every Uα = ∅, in whichcase X \

⋃α∈A Uα = X; (3) X \

⋂α∈A finite Uα =

⋃α∈A finite(X \ Uα) is countable

since it is the finite union of countable sets, unless every Uα = ∅, in which caseX \

⋂α∈A finite Uα = X.

Now consider T∞. It is not a topology, for if we let X = [−1, 1] ⊆ R, and letU1 = [−1, 0) and U2 = (0, 1], we see that both U1, U2 ∈ T∞, but X \ (U1 ∪U2) = 0,which is not infinite, and so U1 ∪ U2 /∈ T∞.

3

Exercise 13.5. Show that if A is a basis for a topology on X, then the topologygenerated by A equals the intersection of all topologies on X that contain A. Provethe same if A is a subbasis.

Proof. Let TA be the topology generated by A, and TI be the intersection of alltopologies that contain A.TI ⊆ TA. This follows from the fact that TA ⊇ A, and so is one of the topologies

that is intersected over in the construction of TI .TA ⊆ TI . Let U ∈ TA; by Lemma 13.1, U =

⋃αAα for some collection Aαα ⊆

A. But U =⋃αAα ∈ TI since each Aα ∈ TI .

Now let A be a subbasis. The proof that TI ⊆ TA is identical; it remains toshow TA ⊆ TI . Let U ∈ TA; by definition of the topology generated by A, U is theunion of a finite intersection of elements Aαα ⊆ A. But then U ∈ TI since eachAα ∈ TI .Exercise 13.6. Show that the topologies of R` and RK are not comparable.

Proof. R` 6⊆ RK . For [a, b) ∈ R`, there is no basis element U ∈ RK such thata ∈ U,U ⊆ [a, b).

RK 6⊆ R`. For (−1, 1) \ K ∈ RK which contains 0, there is no basis element[a, b) ∈ R` such that 0 ∈ [a, b), [a, b) ⊆ (−1, 1) \ K by the Archimedean property,that is, for all ε > 0, there exists N ∈ N such that 1/N < ε.

Exercise 13.7. Consider the following topologies on R:

T1 = the standard topology,

T2 = the topology of RK,

T3 = the finite complement topology,

T4 = the upper limit topology, having all sets (a, b] as basis,

T5 = the topology having all sets (−∞, a) = x | x < a as basis.

Determine, for each of these topologies, which of the others it contains.

Proof. We claim we have the following Hasse diagram:

T1

T3 T5

T2

T4

4

T3 ( T1. Inclusion is true since U ∈ T3 =⇒ U c finite, and so if we let U c =xini=1 with xi in increasing order, U =

⋃ni=0(xi, xi+1), where x0 = −∞, xn+1 =∞.

Inequality follows since for (a, b) such that −∞ < a, b <∞, R \ (a, b) is not finite.T5 ( T1. Inclusion is clear since (−∞, a) is of the form (b, c). Inequality follows

since for (b, c) ∈ T1 and x ∈ (b, c), there is no basis element (−∞, a) ∈ T5 such thatx ∈ (−∞, a), (−∞, a) ⊆ (b, c) (if b > −∞).T3 and T5 are not comparable. T3 6⊆ T5 since R \ 0 ∈ T3, but if we take x > 0,

which is in this set, there is no basis element (−∞, a) ∈ T5 that contains x but iscontained in R \ 0. T5 6⊆ T3 since (−∞, 0)c is not finite.T1 ( T2 by Lemma 13.4.T2 ( T4. For (a, b) ∈ T2 and x ∈ (a, b), (a, x] ∈ T4 and (a, x] ⊆ (a, b). For

(a, b) \K ∈ T2 and x ∈ (a, b) \K, we note that x ∈ (1/(n+ 1), c] where x < c < 1/n,x ∈ (a, 0], or x ∈ (1, d], where x < d < b; in all three cases, these sets are subsets of(a, b) \K and are members of T4. Inequality follows since for (a, b] ∈ T4, there is nobasis element U ∈ T2 such that b ∈ U,U ⊂ (a, b].

16 The Subspace Topology

Exercise 16.8. If L is a straight line in the plane, describe the topology L inheritsas a subspace of R` × R and as a subspace of R` × R`. In each case it is a familiartopology.

Solution. Note that the basis for R`×R consists of elements of the form [a, b)×(c, d).If L = (x, y) | x = x0, then L∩[a, b)×(c, d) = ∅ or x0×(c, d), and so defining themap ϕ : L∩(R`×R)→ R, x0×(c, d) 7→ (c, d), it is bijective, open, and continuous,and so the topology L inherits is homeomorphic to R with the standard topology. IfL has finite slope, we first note that L ∩ (R` × R) = (x,mx + b) ∈ R2 | x ∈ R,and that the basis for our topology are the sets of the form ∅, [(a,ma + b), (c,mc +b)), ((a,ma+ b), (c,mc+ b)) for a, c ∈ R and a < c, by Lemma 16.1. We then define

ϕ : L ∩ (R` × R)→ R`, (a,ma+ b) 7→ a.

This implies

((a,ma+ b), (c,mc+ b)) 7→ (a, c),

[(a,ma+ b), (c,mc+ b)) 7→ [a, c).

We claim this defines a homeomorphism with R`. Clearly, it is continuous, for thebasis elements of R` have preimages that are basis elements in the topology on L.Likewise, it is open since the basis elements of L map to sets that are open in R` by

5

Lemma 13.4. Finally this is a bijection since there exists an inverse just by reversingthe arrows above.

For R` × R`, by following the same steps as above if L = (x, y) | x = x0, thenL∩ (R` ×R`) is homeomorphic to R`. For L with |m| <∞, we must split it up intotwo cases. When m ≥ 0, we have a similar situation as above, except we only haveto consider basis elements of the form [a, b); thus, L ∩ (R` × R`) is homeomorphicto R`. When m < 0, since for every point (x, y) ∈ L, we can find a basis element[x, a)× [y, b) ∈ (R` × R`) such that L ∩ [x, a)× [y, b) = (x, y), and these form theopen sets of our new topology by Lemma 16.1. We see then that the topology on Lis homeomorphic to the discrete topology on R.

Exercise 16.9. Show that the dictionary order topology on the set R×R is the sameas the product topology Rd×R, where Rd denotes R in the discrete topology. Comparethis topology with the standard topology on R2.

Proof. We see that the basis elements of (R × R)lex consist of intervals of the form(a × b, c × d) for a < c, and for a = c and b < d, as in Example 14.2. These basiselements are open in Rd × R since

(a× b, c× d) = (a, c)× R ∪ a × (c,∞) ∪ b × (−∞, d) ∈ TRd×R.

For the reverse situation, consider the basis elements for Rd × R; these consist ofall a × (b, c) since a | a ∈ R forms a basis for Rd by Example 13.3. But then,a×(b, c) are open in R×R with the order topology since it is of the form (a×b, c×d)for a = c.

We now compare this to the standard topology on R2. Since (a, b)×(c, d) ∈ Rd×R,we see that R2 ⊆ Rd × R. Moreover, since a × (b, c) ∈ (Rd × R) \ R2, we see thatR2 ( Rd × R.

17 Closed Sets and Limit Points

Exercise 17.2. Show that if A is closed in Y and Y is closed in X, then A is closedin X.

Proof. A is closed in Y iff there exists B ⊆ X closed in X such that A = Y ∩ B byTheorem 17.2. But then, A is the intersection of closed sets, and so is closed.

Exercise 17.3. Show that if A is closed in X and B is closed in Y , then A× B isclosed in X × Y .

6

Proof. We see that X \A, Y \B are open in X, Y respectively by definition of a closedset. By definition of the product topology, (X\A)×Y,X×(Y \B) are open in X×Y .We see that (X \ A)× Y = (X × Y ) \ (A× Y ), X × (Y \B) = (X × Y ) \ (X ×B),and so A× Y,X ×B are closed in X × Y . Finally, A×B = (A× Y )∩ (X ×B), andso is the intersection of closed sets, i.e., closed.

Exercise 17.5. Let X be an ordered set in the order topology. Show that (a, b) ⊂[a, b]. Under what conditions does equality hold?

Proof. Since (a, b) ⊆ [a, b] closed, and by the definition of closure,

(a, b) =⋂

K⊇(a,b) closed

K ⊆ [a, b].

(a, b) = [a, b] ⇐⇒ a, b ∈ (a, b) ⇐⇒ any basis elements A 3 a,B 3 b intersect(a, b) by Theorem 17.5(b). We claim that this is equivalent to the fact that there isno immediate successor α of a and no immediate predecessor β of b. If either arethe case, say for a, then choosing A with upper bound α would not intersect (a, b),and so equality doesn’t hold since a /∈ (a, b); in the other direction, if neither are thecase, we see that, say for a, the upper bound of A, α would be such that (a, α) isnon-empty, and so A ∩ (a, b) 6= ∅, satisfying the condition for Theorem 17.5(b). Thesame argument applies when considering b and β, and so our claim holds.

Exercise 17.13. Show that X is Hausdorff if and only if the diagonal ∆ = x×x |x ∈ X is closed in X ×X.

Proof. Suppose ∆ is closed in X × X, i.e., the complement ∆c is open. This isequivalent to for all (x, y) ∈ X × X such that x 6= y, there exists a basis elementU × V of X ×X for U, V open in X such that (x, y) ∈ U × V but (U × V )∩∆ = ∅.But then, by definition of ∆, this is equivalent to saying for all x, y ∈ X such thatx 6= y, there exist open neighborhoods U 3 x and V 3 y such that U ∩ V = ∅, andso X is Hausdorff.

Exercise 17.16. Consider the five topologies on R given in Exercise 7 of §13.(a) Determine the closure of the set K = 1/n | n ∈ Z+ under each of these

topologies.(b) Which of these topologies satisfy the Hausdorff axiom? the T1 axiom?

Solution for (a). For T3, A closed ⇐⇒ A finite or all of R. Since no finite setcontains all of K, we see that R is the only closed set containing K, and so K = R.

7

For T5, we claim K = [0,∞). For x ∈ [0,∞), the basis elements that containx are of the form (−∞, a) for a > x. Since (−∞, a) ∩ K 6= ∅ by the Archimedeanproperty, that is, ∀ε > 0∃N ∈ N such that 1/N < ε, K = [0,∞) by Theorem 17.5.

For T1, K ′ = 0 by Example 17.8, and so K = K ∪ 0.For T2, K is closed since R\K = (−∞,∞)\K is a basis element, and so K = K.For T4, K = K since T4 is finer than T2, and so R \K is still open.

Solution for (b). T3 is T1 since any finite point set is closed by definition of the finitecomplement topology. It is not Hausdorff, for if we choose U 3 x, V 3 y both open,(U ∩V )c = U c∪V c is finite, where the equality follows from De Morgan’s Laws, andso U ∩ V is infinite.T5 is not Hausdorff and not even T1, for R\x0 is not a union of basis elements,

and so x0 is not closed.T1 is Hausdorff, for if we have x, y ∈ R and 0 < ε < |x − y|/2, then (x − ε, x +

ε) ∩ (y − ε, y + ε) = ∅. Since Hausdorff =⇒ T1, we see that T1 is also T1.Since T2, T4 are both finer than T1, we see that the open sets constructed above

are still open and separate x, y, and so T2, T4 are still both Hausdorff and thus T1.

18 Continuous Functions

Exercise 18.1. Prove that for functions f : R→ R, the ε-δ definition of continuityimplies the open set definition.

Remark. Recall that f is continuous if for every ε > 0 and x0 ∈ R, there exists aδ > 0 such that |f(x)− f(x0)| < ε for all x ∈ R such that |x− x0| < δ.

Proof. Consider x0 ∈ R, and a corresponding neighborhood V of f(x0); we thenhave V ⊇ (f(x0)− ε, f(x0) + ε) for some ε > 0 since V is open. Then, by hypothesisthere exists a δ > 0 such that f(x) ∈ (f(x0) − ε, f(x0) + ε) for all x ∈ R such thatx ∈ (x0 − δ, x0 + δ) = U , which is open. Thus, f(U) ⊆ V , and so f is continuous byTheorem 18.1.

Exercise 18.12. Let F : R× R→ R be defined by the equation

F (x× y) =

xy/(x2 + y2) if x× y 6= 0× 0.

0 if x× y = 0× 0.

(a) Show that F is continuous in each variable separately.(b) Compute the function g : R→ R defined by g(x) = F (x× x).(c) Show that F is not continuous.

8

Proof of (a). Since F is symmetric in interchanging x ↔ y, we only have to prove∀y0 ∈ Y , h(x) = F (x × y0) is continuous as a function R → R. For y0 = 0, this istrivially true for the image of h is (0, 0) with preimage R. Now suppose y0 6= 0; thenwe have h(x) = xy0/(x

2 + y20). This is continuous since xy0 and x2 + y2

0 are bothcontinuous, and so their quotient is also continuous (since also x2 + y2

0 6= 0), usingthe ε-δ definition of continuity (see Theorem 21.5).

Proof of (b). Since F (x× x) for x 6= 0 equals x2/(x2 + x2) = x2/2x2 = 1/2, we have

g(x) =

1/2 if x 6= 0.

0 if x = 0.

Proof of (c). We claim F (x×y) is not continuous along the line L = x = y at (0, 0),i.e., F |L is not continuous at (0, 0); this suffices by Theorem 18.2(d). Note that theline L in the subspace topology is homeomorphic to R, where the homeomorphismis given by either of the coordinate projection maps. Now the preimage of the closedset 1/2 ⊆ R is L \ (0, 0), which is not closed since R \ 0 is not closed, henceF |L is not continuous, and neither is F .

19 The Product Topology

Exercise 19.6. Let x1,x2, . . . be a sequence of the points of the product space∏Xα.

Show that this sequence converges to the point x if and only if the sequence πα(x1),πα(x2), . . . converges to πα(x) for each α. Is this fact true if one uses the box topologyinstead of the product topology?

Proof. Suppose xi → x, and fix some index γ. Then, for any neighborhood Uγ 3πγ(x), letting U =

∏Uα where Uα = Xα for all α 6= γ, there exists N ∈ N such that

xi ∈ U for all i ≥ N , and so πγ(xi) ∈ πγ(U) = Uγ for all i ≥ N , i.e., πγ(xi) →πγ(x). Note that this direction does not depend on the topology being the productor box topology.

In the other direction, suppose πα(xi) → πα(x) for all α. We take an arbitraryneighborhood V of x ∈

∏Xα; it then contains a basis element of

∏Xα containing

x, which is a product of open sets∏Uα. In the case of the product topology, there

then exist only finite Uα ( Xα, and for these open sets there exist Nα ∈ N such thatπα(xi) ∈ Uα for all i ≥ Nα for each α. Nα = 1 works for all other α. Thus, we cantake N = max(Nα); then, xi ∈

∏Uα ⊆ V for all i ≥ N .

We construct a counterexample for this direction in the case of the box topology.Let RN be the box topology on the product of copies of R indexed by N, and let

xi := (1i, 1i, 1i, . . .).

9

Then, for each α ∈ N, πα(xi) → (0, 0, 0, . . .) =: x, but this sequence does notconverge in the box topology, for the open set∏

i∈N

(−1i, 1i) = (−1, 1)× (−1

2, 1

2)× (−1

3, 1

3)× · · ·

in the box topology contains x = (0, 0, 0, . . .), but does not contain any xi.

20 The Metric Topology

Exercise 20.4. Consider the product, uniform, and box topologies on Rω.(a) In which topologies are the following functions from R to Rω continuous?

f(t) = (t, 2t, 3t, . . .),

g(t) = (t, t, t, . . .),

h(t) = (t, 12t, 1

3t, . . .).

(b) In which topologies do the following sequences converge?

w1 = (1, 1, 1, 1, . . .), x1 = (1, 1, 1, 1, . . .),

w2 = (0, 2, 2, 2, . . .), x2 = (0, 12, 1

2, 1

2, . . .),

w3 = (0, 0, 3, 3, . . .), x3 = (0, 0, 13, 1

3, . . .),

. . . . . .

y1 = (1, 0, 0, 0, . . .), z1 = (1, 1, 0, 0, . . .),

y2 = (12, 1

2, 0, 0, . . .), z2 = (1

2, 1

2, 0, 0, . . .),

y3 = (13, 1

3, 1

3, 0, . . .), z3 = (1

3, 1

3, 0, 0, . . .),

. . . . . .

Solution for (a). For the product topology, by Theorem 19.6, f, g, h are all continu-ous since each coordinate function is continuous. This is because if an open set inthe image of a coordinate function is (a, b), its preimage would still be in the form(a′, b′) ⊆ R where a′, b′ are determined by the linear equations defining f, g, h above.

Now consider the uniform topology. Note by Theorem 21.1 we can use the familiarε-δ definition for continuity since our spaces both are metric spaces. We claim f is notcontinuous. For, suppose it is continuous. Then, given ε > 0 and x ∈ R, there existsδ > 0 such that |x−y| < δ =⇒ |f(x)−f(y)| = supn[min(n|x−y|, 1)] < ε. But, this isa contradiction since for n large, min(n|x−y|, 1) = 1, and so is always greater than ε.Now consider g. g is continuous since given ε > 0 and x ∈ R, we let δ < min(ε, 1) and

10

therefore have |x−y| < δ =⇒ |f(x)−f(y)| = supn[min(|x−y|, 1)] = min(|x−y|, 1) <min(ε, 1) ≤ ε. h is also continuous since given ε > 0 and x ∈ R, we let δ < min(ε, 1)and therefore have |x − y| < δ =⇒ |f(x) − f(y)| = supn[min(|x − y|/n, 1)] ≤min(|x− y|, 1) < min(ε, 1) ≤ ε.

For the box topology, since the box topology is finer than the uniform topologyby Theorem 20.4, we see that f is not continuous. For, if V open in the uniformtopology has preimage that is not open in R, V is still open in the box topology andstill has the same non-open preimage. Next, by Example 19.2, we see that g is notcontinuous. Last, for h, we choose

B = (−1, 1)× (− 122, 1

22)× (− 1

32, 1

32)× · · · ,

and suppose its preimage h−1(B) is open. This implies h((−δ, δ)) ⊆ B, and soapplying πn gives

hn((−δ, δ)) = (− δn, δn) ⊆ (− 1

n2 ,1n2 )

for all n, a contradiction.

Solution for (b). We note that since the product topology is Hausdorff by Theorem19.4 and both the uniform and box topologies are finer than the product topology byTheorems 19.1 and 20.4, if a sequence converges to a point p in one topology, it mustconverge to the same point in the finer topologies. For, if the sequence converges toq in the finer topology, then it also converges to q in the coarser topology, and bythe Hausdorff property p = q.

Consider wn. For the product topology, we recall that any basic open set U =∏Uα 3 0 is the product of finitely many open subsets of R with infinitely many

copies of R. Letting N be the largest α such that Uα ( R, we see that wn ∈ Ufor all n > N since the first N components are zero, and the rest are trivially inthe remaining copies of R of U . Thus, wn → 0 in the product topology. Now weonly have to check if the sequence converges to zero in the other topologies by theabove. In the uniform topology, ρ(wn, 0) = 1 for all n, and so the sequence does notconverge. For, if we choose any ball U = B(0, r) ⊆ Rω for r < 1, wn /∈ U for all n.Finally, since the box topology is finer than the uniform topology by Theorem 20.4,we see that this same open set U is such that wn /∈ U for all n, and so wn does notconverge in the box topology, either.

Consider xn and yn. We claim they both converge to zero in the uniform topology.For any open set 0 ∈ U ⊆ Rω in the uniform topology, we can find B(0, ε) such thatB(0, ε) ⊆ U ; then, we can find N such that 1/N < ε. We then see that xn,yn ∈B(0, ε) ⊆ U for all n ≥ N , and so xn,yn → 0 in the uniform topology. Moreover,since the uniform topology is finer than the product topology, we see that this implies

11

xn,yn → 0 in the product topology as well. For the box topology, though, we see thatneither sequence converges. For, we can construct the set 0 ∈ U =

∏∞n=1(−1/n, 1/n)

(where we only consider sets containing zero by the above), which does not containxn,yn for any n.

For zn, we see that for any open set 0 ∈ U =∏Uα ⊆ Rω in the box topology,

for N large enough 1/n ∈ U1, U2 for n ≥ N , and so zn ∈ U for all n ≥ N , since byhypothesis 0 ∈ U , the third component onwards of zn are always in their respectiveUα. Thus, zn → 0 in the box topology; since the box topology is finer than both theuniform and product topologies, we see that this implies zn converges in the othertwo topologies as well.

Exercise 20.5. Let R∞ be the subset of Rω consisting of all sequences that areeventually zero. What is the closure of R∞ in Rω in the uniform topology? Justifyyour answer.

Solution. We claim that A = R∞ is the set of all sequences that converge to zero; wedenote this latter set by X. It suffices to show by Theorem 17.5 that x ∈ X if andonly if every basis element U 3 x intersects A. First suppose x ∈ X and let U 3 xbe a basis element in the uniform topology; we then see that we can find an openball B(x, ε) ⊆ U . We know we can find N ∈ N such that |xn| < ε for all n ≥ N bythe definition of convergence. Then, define y such that yn = xn for all n < N , andzero otherwise; this means y ∈ A. Then, ρ(x, y) < ε, and so y ∈ B(x, ε) ∩ A.

Now suppose x /∈ X; it suffices to find a basis element containing x that does notintersect A. Since x /∈ X, there exists a ball B(0, ε) ⊆ R such that xnn≥N 6⊆ B(0, ε)for any N . The ball B(x, ε/2), then, does not intersect A, since for any y ∈ B(x, ε/2),it is not the case that yn = 0 for all n ≥ N for some N .

Exercise 20.6. Let ρ be the uniform metric on Rω. Given x = (x1, x2, . . .) ∈ Rω

and given 0 < ε < 1, let

U(x, ε) = (x1 − ε, x1 + ε)× · · · × (xn − ε, xn + ε)× · · · .

(a) Show that U(x, ε) is not equal to the ε-ball Bρ(x, ε).(b) Show that U(x, ε) is not even open in the uniform topology.(c) Show that

Bρ(x, ε) =⋃δ<ε

U(x, δ).

Proof of (a). Consider the point

y = (yn)n∈N, yn = xn + ε

n∑k=1

1

2k= xn + ε

(1− 1

2n

).

12

yn ∈ (xn − ε, xn + ε) for all n implies y ∈ U(x, ε), while y /∈ Bρ(x, ε) since

ρ(x,y) = supnd(xn, yn) = ε.

Proof of (b). U(x, ε) is not open since the point y in (a) has no neighborhood con-tained in Bρ(x, ε). For, suppose Bρ(y, δ) ⊆ U(x, ε). We can find N such that

δ

2> ε

∞∑k=N+1

1

2k,

since∑

1/2k converges, and so its tail becomes infinitesimally small. We see thatthen, defining y′ such that y′n = yn for all n 6= N and y′N = yN + δ/2, y′ ∈ Bρ(y, δ)but y′ /∈ U(x, ε) since y′n = yn+δ/2 > yn+ε

∑∞k=N+1

12k

= xn+ε, a contradiction.

Proof of (c). The ⊇ direction is clear, since each U(x, δ) ⊆ Bρ(x, ε) by the fact thatδ < ε. Now suppose z ∈ Bρ(x, ε); if ρ(x, z) = ξ, then we can find δ ∈ (ξ, ε) so thatz ∈ U(x, ξ), i.e., Bρ(x, ε) ⊆

⋃δ<ε U(x, δ).

Exercise 20.8. Let X be the subset of Rω consisting of all sequences x such that∑x2i converges. Then the formula

d(x,y) =

[∞∑i=1

(xi − yi)2

]1/2

defines a metric on X. On X we have the three topologies it inherits from the box,uniform, and product topologies on Rω. We have also the topology given by the metricd, which we call the `2-topology.

(a) Show that on X, we have the inclusionsbox topology ⊃ `2-topology ⊃ uniform topology.

(b) The set R∞ of all sequences that are eventually zero is contained in X. Showthat the four topologies that R∞ inherits as a subspace of X are all distinct.

(c) The set

H =∏n∈Z+

[0, 1/n]

is contained in X; it is called the Hilbert cube. Compare the four topologiesthat H inherits as a subspace of X.

13

Proof of (a). box topology ⊃ `2-topology. Let U 3 x be a basis element in the `2-topology. Then, there exists a basis element U = Bd(x, ε) ⊆ U of the `2-topology.We claim that V = X ∩

∏(xi − ε/2i/2, xi + ε/2i/2) 3 x basis element of the box

topology is contained in U . Suppose y ∈ V ; then

d(x,y)2 =∞∑i=1

(xi − yi)2 <

∞∑i=1

ε2

2i= ε2 =⇒ y ∈ U,

i.e., V ⊆ U ⊆ U , and box topology ⊃ `2-topology by Lemma 13.3.`2-topology ⊃ uniform topology. Let U 3 x be a basis element in the uniform

topology. Then, there exists a basis element U = X ∩ Bρ(x, ε) ⊆ U of the uniformtopology. We claim V = Bd(x, ε) basis element of the `2-topology is contained in U .If ε > 1, then trivially V ⊆ U = X, so we assume ε ≤ 1. Suppose y ∈ V ; then

ρ(x,y) = sup |xi − yi| ≤ d(x,y) < ε =⇒ y ∈ U,

i.e., V ⊆ U ⊆ U , and `2-topology ⊃ uniform topology by Lemma 13.3.

Proof of (b). By (a) and Theorem 20.4, we have the inclusions

box topology ⊃ `2-topology ⊃ uniform topology ⊃ product topology,

since if T1 ⊃ T2 in the ambient space, we would also have T ′1 ⊃ T ′2 , where T ′i arethe subspace topologies induced by Ti on X, by the fact that the open sets in thesubspace X are the open sets of the ambient space intersected with X. We claimthese are strict inclusions.

box topology ) `2-topology. Consider the open set U = R∞ ∩∏

(−1/i, 1/i) 3 0in the box topology. Consider any neighborhood V 3 0; then, there exists someV = R∞ ∩ Bd(0, ε) for ε > 0 contained in V . We can find N ∈ N such that1/N < ε, and so x such that xi = 0 for all i except xN = 1/N is contained in Vsince d(0,x) = 1/N < ε and therefore V , but not in U . Hence, U is open in the boxtopology but not in the `2-topology by p. 78, and so box topology ) `2-topology.

`2-topology ) uniform topology. Consider the open set U = R∞ ∩ Bd(0, 1) inthe `2-topology. Consider any neighborhood V 3 0; then, there exists some V =R∞ ∩ Bρ(0, ε) for ε > 0 contained in V . We can find N ∈ N such that Nε2 > 4, andso x such that xi = ε/2 for all 1 ≤ i ≤ N and 0 otherwise is contained in V andtherefore V , yet

d(x, 0)2 =∞∑i=1

x2i = N

ε2

4> 1 =⇒ d(x, 0) > 1,

14

and so x /∈ U . Hence, U is open in the `2-topology but not in the uniform topologyby p. 78, and so `2-topology ) uniform topology.

uniform topology ) product topology. Consider the open set U = R∞ ∩Bρ(0, 1)in the uniform topology. Consider any V = R∞∩

∏Uα 3 0 where Uα = R for all but

finitely many α. Let N be such that UN = R; then, x such that xi = 0 for all i except|xN | ≥ 1 is in V but not in U . Hence, U is open in the uniform topology but not inthe product topology by p. 78, and so uniform topology ) product topology.

Solution for (c). We claim that

box topology ) (`2-topology = uniform topology = product topology),

i.e., the box topology is strictly finer than the other topologies, which are equal.To show the equality, it suffices to show product topology ⊃ `2-topology, for

then we have `2-topology ⊃ uniform topology ⊃ product topology ⊃ `2-topology bythe same argument as in (b), and so we must have equality throughout. So, considerU 3 x open in the `2-topology; then, we can find a basis element U = H∩Bd(x, ε) ⊆ Uof the `2-topology for some ε > 0. Let δ = ε/[ζ(2) + 1]1/2, and choose N such that∑∞

i=N 1/i2 < δ2, which exists since |ζ(2)| <∞. We claim that

V = H ∩

[N−1∏i=1

(xi −

δ

i, xi +

δ

i

)×∞∏i=N

R

]3 x,

basis element of the product topology is contained in U . Suppose y ∈ V ; then

d(x,y)2 =∞∑i=1

(xi − yi)2 < δ2

N∑i=1

1

i2+∞∑i=N

1

i2< δ2[ζ(2) + 1] = ε2 =⇒ y ∈ U,

i.e., V ⊆ U ⊆ U , and product topology ⊃ `2-topology by Lemma 13.3. Thus, wehave the equality desired.

It remains to show the box topology is strictly finer than the other topologies;since the other topologies are equal it suffices to show box topology ) `2-topology.But the open set U = H ∩

∏(−1/i, 1/i) 3 0 is open in the box topology but not the

`2-topology by the same argument as the proof that box topology ) `2-topology in(b), and so we are done.

21 The Metric Topology (continued)

Exercise 21.1. Let A ⊆ X. If d is a metric for the topology of X, show that d|A×Ais a metric for the subspace topology on A.

15

Proof. Clearly d|A×A is a metric since it inherits all the properties for a metric fromthe metric d for X; it therefore suffices to show that every basis element for thesubspace topology on A contains some open ball defined by d|A×A, and vice versa,by Lemma 13.3.

So, suppose B is a basis element for the metric topology on A; B = Bd|A×A(x, r)

for some x ∈ A and r ∈ R. But then, B = Bd(x, r) ∩ A by definition, and so thesubspace topology is finer than the metric topology.

Conversely, suppose B is a basis element for the subspace topology on A; it equalsBd(x, r)∩A for some basis element Bd(x, r) of X. Let y ∈ Bd(x, r)∩A; we see thatthe set Bd|A×A

(y, r−d(x, y)) ⊆ A∩Bd(x, r) is a basis element for the metric topologycontained in A ∩Bd(x, r), since z ∈ Bd|A×A

(y, r − d(x, y)) is such that

d(z, x) ≤ d(z, y) + d(x, y) = d|A×A(z, y) + d(x, y) < r − d(x, y) + d(x, y) = r.

Thus, the metric topology is finer than the subspace topology. Combining the twoinclusions, we see the topologies are equal.

Exercise 21.2. Let X and Y be metric spaces with metrics dX and dY , respectively.Let f : X → Y have the property that for every pair of points x1, x2 of X,

dY (f(x1), f(x2)) = dX(x1, x2).

Show that f is an imbedding. It is called an isometric imbedding of X in Y .

Proof. We first show f is injective:

f(x1) = f(x2) =⇒ dY (f(x1), f(x2)) = 0 =⇒ dX(x1, x2) = 0 =⇒ x1 = x2

by properties of metrics, and so we have an injective map.Now we consider the map f ′ : X → im(X) ⊆ Y , which is a bijection; it suffices to

show that f ′, f ′−1 are continuous to show f is an imbedding. Let x ∈ X and ε > 0be given; then, letting δ = ε, we have

dX(x, y) < δ =⇒ dY (f ′(x), f ′(y)) < ε,

and so f is continuous. Given y ∈ Y and ε > 0, letting δ = ε gives

dY (x, y) = dY (f(a), f(b)) < δ =⇒ dX(f ′−1(x), f ′−1(y)) = dX(a, b) < ε,

where a, b exist by the bijectivity of f , and so f ′−1 is continuous.

Exercise 21.3. Let Xn be a metric space with metric dn, for n ∈ Z+.

16

(a) Show thatρ(x, y) = maxd1(x1, y1), . . . , dn(xn, yn)

is a metric for the product space X1 × · · · ×Xn.(b) Let di = mindi, 1. Show that

D(x, y) = supdi(xi, yi)/i

is a metric for the product space∏Xi.

Proof of (a). ρ satisfies properties (1) and (2) on p. 119 since the components do;it then suffices to show the triangle inequality. We first have di(xi, zi) ≤ d(xi, yi) +d(yi, zi) for all i. Then, by definition of ρ, di(xi, zi) ≤ ρ(x, y) + ρ(y, z) for all i. Butsince this is true for all i, we have that ρ(x, z) ≤ ρ(x, y) + ρ(y, z).

We now show that this defines a metric for the product space. First let B =∏Ui

be a basis element of∏Xi, and let x ∈ B. For each i, there is an εi such that

Bdi(xi, εi) ⊆ Ui. Choosing ε = minε1, . . . , εn, we see that Bρ(x, ε) ⊆ B, since ify ∈ Bρ(x, ε), di(xi, yi) ≤ ρ(x,y) < ε ≤ εi, and so y ∈

∏Ui as desired. Thus the

metric topology is finer than the product topology.Conversely, let Bρ(x, ε) be a basis element in the metric topology; since it is the

product Bdi(xi, ε), we see that the product topology is finer than the metric topology.These two facts imply the two topologies are equal.

Proof of (b). D satisfies properties (1) and (2) on p. 119 since the components do; itthen suffices to show the triangle inequality. We first have

di(xi, zi)

i≤ di(xi, yi)

i+di(yi, zi)

i≤ D(x, y) +D(y, z)

for all i. But since this is true for all i, we have that

D(x, z) = sup

di(xi, zi)

i

≤ D(x, y) +D(y, z).

We now show that this defines a metric for the product space. Let U be open inthe metric topology and let x ∈ U ; choose an open ball BD(x, ε) ⊆ U . Choose Nsuch that 1/N < ε, and let

V = Bd1(x1, ε)× · · · ×BdN

(xN , ε)×XN+1 ×XN+2 × · · · .

We claim V ⊆ BD(x, ε) ⊆ U . Given y ∈∏Xi, di(xi, yi)/i ≤ 1/N for i ≥ N .

Therefore,

D(x,y) ≤ max

d1(x1, z1)

1, · · · , dN(xN , zN)

N,

1

N

.

17

If y ∈ V , this expression is less than ε, so V ⊆ BD(x, ε) as desired, and the producttopology is finer than the metric topology.

Conversely, let U =∏Ui where Ui is open in Xi for α1, . . . , αn and Ui = Xi

otherwise. Let x ∈ U be given, and choose Bdi(xi, εi) ⊆ Xi for i = α1, . . . , αn,

where each εi ≤ 1. Then, defining ε = minεi/i | i = α1, . . . , αn, we claim thatx ∈ BD(x, ε) ⊆ U . Let y be a point of BD(x, ε). Then, for all i,

di(xi, yi)

i≤ D(x,y) < ε.

Now if i = α1, . . . , αn, then ε ≤ εi/i, so that di(xi, yi) < εi ≤ 1; it follows that|xi − yi| < εi, and so y ∈

∏Ui as desired. We thus have that the metric topology is

finer than the product topology; combined with the above this implies the topologiesare equal.

22 The Quotient Topology

Exercise 22.2.(a) Let p : X → Y be a continuous map. If there is a continuous map f : Y → X

such that p f equals the identity map of Y , then p is a quotient map.(b) If A ⊂ X, a retraction of X onto A is a continuous map r : X → A such

that r(a) = a for each a ∈ A. Show that a retraction is a quotient map.

Proof of (a). If V ⊆ Y with U = p−1(V ) ⊆ X open, f−1(U) = f−1(p−1(V )) =(p f)−1(V ) = V is open. Thus, p is a quotient map.

Proof of (b). Let ι : A → X be the inclusion map; then, r ι is the identity on A,hence r is a quotient map by (a).

Exercise 22.4.(a) Define an equivalence relation on the plane X = R2 as follows:

x0 × y0 ∼ x1 × y1 if x0 + y20 = x1 + y2

1.

Let X∗ be the corresponding quotient space. It is homeomorphic to a familiarspace; what is it?

(b) Repeat (a) for the equivalence relation

x0 × y0 ∼ x1 × y1 if x20 + y2

0 = x21 + y2

1.

18

Solution for (a). Set g(x × y) = x + y2 ∈ R. We see it is a surjection onto Rsince R × 0 7→ R. It is continuous since for x0 × y0 ∈ X, given ε > 0, lettingδ = min(1, ε/2(|y0|+ 1)), ρ(x0 × y0, x× y) < δ implies

|g(x0 × y0)− g(x× y)| = |(x0 + y20)− (x+ y2)|

≤ |x0 − x|+ |y20 − y2|

≤ |x0 − x|+ |y0 + y||y0 − y|≤ |x0 − x|+ |y − y0|2 + 2|y0||y0 − y|< 2(|y0|+ 1)δ < ε.

If we define f : R → X by x 7→ x× 0, which is continuous since (a, b)× (c, d) mapsback to (a, b) which is open in R, we see gf is the identity on R, and so g is a quotientmap by the lemma above. Since x0 × y0 ∼ x1 × y1 ⇐⇒ g(x0 × y0) = g(x1 × y1),by Corollary 22.3, this induces a bijective continuous map g′ : X∗ → R, which is ahomeomorphism since g was a quotient map.

Solution for (b). Set g(x×y) = x2 +y2 ∈ R. We see it is a surjection onto R≥0, sinceR×0 7→ R≥0, and it does not map to anything else since x2 +y2 ≥ 0 for all x, y. Itis continuous since for x0×y0 ∈ X, given ε > 0, letting δ = min(1, ε/2(|y0|+|x0|+1)),ρ(x0 × y0, x× y) < δ implies

|g(x0 × y0)− g(x× y)| = |(x20 + y2

0)− (x2 + y2)|≤ |x2

0 − x2|+ |y20 − y2|

≤ |x0 − x||x− x0 + 2x0|+ |y0 − y||y − y0 + 2y0|≤ |x0 − x|(1 + 2|x0|) + |y0 − y|(1 + 2|y0|)≤ 2δ(|x0|+ |y0|+ 1) < ε.

We define f : R≥0 → X by x 7→√x × 0, which is continuous since the preimage of

(a, b) × (c, d), if (c, d) 3 0, is the open set R≥0 ∩ (a′, b′), where a′ = a2 if a ≥ 0,and −1 otherwise, and similarly for b′ (we chose −1 out of convenience; we reallyonly have to make sure the preimage is a half-open set [0, b′) or the empty set inthese cases); if (c, d) 63 0, then the preimage would be empty. We then see g fis the identity on R≥0, and so g is a quotient map by the lemma above. Sincex0 × y0 ∼ x1 × y1 ⇐⇒ g(x0 × y0) = g(x1 × y1), by Corollary 22.3, this induces abijective continuous map g′ : X∗ → R≥0, which is a homeomorphism since g was aquotient map.

Exercise 22.6. Recall that RK denotes the real line in the K-topology. Let Y be thequotient space obtained by RK by collapsing the set K to a point; let p : RK → Y bethe quotient map.

19

(a) Show that Y satisfies the T1 axiom, but is not Hausdorff.(b) Show that p× p : RK × RK → Y × Y is not a quotient map.

Proof of (a). Recall by p. 141 that it suffices to show every element in the partition,i.e., one-point sets x for x /∈ K and K itself, are closed in RK . The former areclosed since RK is T1 since it is Hausdorff by Example 31.1, and the latter is closedsince it is the complement of R \K. Thus, Y is T1.

We now show Y is not Hausdorff. We claim that p(0), p(K) are not separable; notep(0) 6= p(K) since they are in different equivalence classes. Suppose Y is Hausdorff,and let V1 3 p(0), V2 3 p(K) be a separation in Y ; they have open preimagesU1 = p−1(V1) 3 0, U2 = p−1(V2) ⊇ K by definition of a quotient map. There thenexists (a, b) \K 3 0 contained in U1, and choosing n ∈ N such that 1/n < b, thereexists (c, d) 3 1/n contained in U2, where we can assume 1/(n+ 1) ≤ c, since if not,we can take the intersection with (1/(n + 1), d). Then, (c, 1/n) ⊆ U1 ∩ U2, and sop((c, 1/n)) ⊆ V1 ∩ V2, which is a contradiction, and so Y is not Hausdorff.

Proof of (b). By Exercise 17.13, we see that since Y is not Hausdorff by (a), thediagonal ∆Y ⊆ Y × Y is not closed. (p−1 × p−1)(∆Y ) = ∆K ∪ (K × K), where∆K ⊆ RK × RK is the diagonal in RK . However, ∆K is closed by Exercise 17.13since RK is Hausdorff by Example 31.1, and so ∆K ∪ (K × K) is closed since is isthe finite union of closed sets. Thus, the inverse image of the non-closed set ∆Y isclosed, and so p× p is not a quotient map.

3 Connectedness and Compactness

23 Connected Spaces

Exercise 23.8. Determine whether or not Rω is connected in the uniform topology.

Solution. Let Rω = A∪B, where A is the set of bounded sequences and B is the setof unbounded sequences of reals. A,B are disjoint, and so it remains to show theyare open. Suppose a = (a1, a2, . . .) ∈ A and b = (b1, b2, . . .) ∈ B. Since |ai| < N forall i for some N , and since |bi| > N + 1 for all i larger than some I, we have thatd(ai, bi) = 1 for all i ≥ I. Thus, ρ(a, b) = 1 for any a ∈ A, b ∈ B, and so the openballs with radius 1/2 around a, b are fully contained in A,B respectively.

Exercise 23.11. Let p : X → Y be a quotient map. Show that if each set p−1(y)is connected, and if Y is connected, then X is connected.

20

Proof. Suppose not. Then, X = A ∪ B for A,B open, disjoint sets. ConsiderC = y ∈ Y | p−1(y) ⊆ A, D = y ∈ Y | p−1(y) ⊆ B; we see that these setsare such that C ∪D = Y since p−1(y) connected implies it is in either A or B byLemma 23.2. C,D are then disjoint by definition and p−1(C) = A, p−1(D) = B bythe fact that p is surjective. p quotient map implies that C,D are then open, and soY = C ∪D is a separation, a contradiction.

24 Connected Subspaces of the Real Line

Exercise 24.7.(a) Let X and Y be ordered sets in the order topology. Show that if f : X → Y is

order preserving and surjective, then f is a homeomorphism.(b) Let X = Y = R+. Given a positive integer n, show that the function f(x) = xn

is order preserving and surjective. Conclude that its inverse, the nth rootfunction, is continuous.

(c) Let X be the subspace (−∞,−1)∪[0,∞) of R. Show that the function f : X →R defined by setting f(x) = x + 1 if x < −1, and f(x) = x if x ≥ 0, is orderpreserving and surjective. Is f a homeomorphism? Compare with (a).

Proof of (a). f is injective since if f(a) = f(b) but a 6= b, then (with possibleswapping) a < b, and so f(a) < f(b), a contradiction. We thus must show fand f−1 are continuous. But f is continuous since f−1((a, b)) = (f−1(a), f−1(b)) isopen (apply the same argument to the intervals of the form [a0, b), (a, b0] for a0, b0

minimal and maximal respectively); the same argument applies for f−1 as well.

Proof of (b). f(x) = xn is order preserving since a < b =⇒ a/b < 1 =⇒ an/bn <1 =⇒ an < bn =⇒ f(a) < f(b). f is continuous since it is the product of ncopies of the identity function, which is continuous. We want to show f is surjective.Letting N = xn | x ∈ Z≥0, we see that every real number y ∈ Y is between twoconsecutive members of N , or it is already an nth power of an integer, in which caseit is trivially mapped to by its nth root. In the case y ∈ Y is not an nth power,we have f(n) < y < f(n + 1), and so by the Intermediate value theorem (Theorem24.3), we see that there exists a point c ∈ X such that f(c) = r, i.e., f is surjective.

Since f is order preserving and surjective, by (a) it is then a homeomorphism,and so f−1, the nth root function, is also continuous.

Proof of (c). f is order-preserving on (−∞,−1) since a < b =⇒ f(a) = a + 1 <b + 1 = f(b), and on [0,∞) since it is the identity. We check that it is orderpreserving around the boundary. So, suppose a < −1 and b ≥ 0. Then, a < b but

21

also a+1 < 0 ≤ b, and so f is order-preserving. f is surjective since if x ∈ R, if x ≥ 0its preimage is itself, and if x < 0, its preimage is x− 1. f is not a homeomorphismby Theorem 23.6 since R is connected but X is not, by considering f−1(R).

This does not contradict (a) since X is not in the order topology. Even if R is inthe order topology, the subspace topology induced on X is not the order topology.For, (−1/2, 1) is open in R, and so (−1/2, 1)∩X = [0, 1) is open in X, but not openin the order topology on X.

Exercise 24.8.(a) Is a product of path-connected spaces necessarily path connected?(b) If A ⊂ X and A is path connected, is A necessarily path connected?(c) If f : X → Y is continuous and X is path connected, is f(X) necessarily path

connected?(d) If Aα is a collection of path-connected subspaces of X and if

⋂Aα 6= ∅, is⋃

Aα necessarily path connected?

Solution for (a). Yes. LetX =∏Xα, x,y ∈ X. Since eachXα is path connected, we

have fα : [0, 1]→ Xα continuous such that fα(0) = xα, fα(1) = yα, where we assumethe closed interval is [0, 1] after composition with multiplication and addition, whichare continuous operations. Thus we have the function f = (fα), which is continuousby Theorem 19.6, with f(0) = x, f(1) = y, and so X is path-connected.

Solution for (b). No, since S in Example 24.7 is not path-connected while S is: it isthe image of the continuous map x 7→ (x, sin(1/x)) from R>0 to R2.

Solution for (c). Yes. For, let x, y ∈ f(X), and choose x0 ∈ f−1(x), y0 ∈ f−1(y).Then, there exists continuous g : [a, b] → X such that g(a) = x0, g(b) = y0, and soits composition f g : [a, b]→ Y is continuous with (f g)(a) = x, (f g)(b) = y.

Solution for (d). Yes. Let x, y ∈⋃Aα and p ∈

⋂Aα. Then, there exists a contin-

uous map f : [a, b] →⋃Aα with f(a) = x, f(b) = p, and similarly g : [b, c] →

⋃Aα

with f(b) = p, f(c) = y, since a, p ∈ Aα for some α and similarly for y (we are freeto have Dom g = [b, c] by composition with multiplication and addition, which arecontinuous). Then, by the pasting lemma (Theorem 18.3) since f, g are continuousand f(b) = g(b), we see that h = f on [a, b] and h = g on [b, c] is a continuous mapsuch that h(a) = x, h(c) = y.

Exercise 24.12. Recall that SΩ denotes the minimal uncountable well-ordered set.Let L denote the ordered set SΩ × [0, 1) in the dictionary order, with its smallestelement deleted. The set L is a classical example in topology called the long line.

22

Theorem. The long line is path connected and locally homeomorphic to R, but itcannot be imbedded in R.

(a) Let X be an ordered set; let a < b < c be points of X. Show that [a, c) hasthe order type of [0, 1) if and only if both [a, b) and [b, c) have the order typeof [0, 1).

(b) Let X be an ordered set. Let x0 < x1 < · · · be an increasing sequence of pointsof X; suppose b = supxi. Show that [x0, b) has the order type of [0, 1) if andonly if each interval [xi, xi+1) has the order type of [0, 1).

(c) Let a0 denote the smallest element of SΩ. For each element a of SΩ differentfrom a0, show that the interval [a0 × 0, a× 1) of SΩ × [0, 1) has the order typeof [0, 1).

(d) Show that L is path connected.(e) Show that every point of L has a neighborhood homeomorphic with an open

interval in R.(f) Show that L cannot be imbedded in R, or indeed in Rn for any n.

Proof of (a). We first note order-preserving maps are injective. Letting f : A → Bbe such a map, if a1, a2 ∈ A, then one is larger than the other by the comparabilityproperty of order relations, so one of f(a1), f(a2) is larger than the other, henceunequal.

Now suppose [a, c) has order type [0, 1), and let f : [0, 1) → [a, c) be the orderisomorphism. We claim

g(x) = f[f−1(b)]x, h(x) = ff−1(b) + [1− f−1(b)]x

define order isomorphisms g : [0, 1) → [a, b) and h : [0, 1) → [b, c). They are order-preserving since if x, y ∈ [0, 1),

x < y =⇒ [f−1(b)]x < [f−1(b)]y =⇒ g(x) = f[f−1(b)]x < f[f−1(b)]y = g(y)

x < y =⇒ f−1(b) + [1− f−1(b)]x < f−1(b) + [1− f−1(b)]y

=⇒ h(x) = ff−1(b) + [1− f−1(b)]x < ff−1(b) + [1− f−1(b)]y = h(y)

where the first implications are due to our linear transformations being strictly mono-tonic increasing, and the second since f is order-preserving. This also implies injec-tivity by the above. It remains to show surjectivity. Let z ∈ [a, b), z′ ∈ [b, c). Then,0 ≤ f−1(z) < f−1(b) and f−1(b) ≤ f−1(z′) < 1, and so

g

[f−1(z)

f−1(b)

]= z, h

[f−1(z′)− f−1(b)

1− f−1(b)

]= z′.

23

Conversely, suppose [a, b) and [b, c) have order type [0, 1), and let g : [0, 1)→ [a, b),h : [0, 1)→ [b, c) be the order isomorphisms. We claim

f(x) =

g(2x) if 0 ≤ x < 1/2

h(2x− 1) if 1/2 ≤ x < 1

is an order isomorphism. It preserves orders since g, h preserve orders on theirrespective domains, and since if x < 1/2 ≤ y, applying f gives

f(x) = g(2x) < b ≤ h(2y − 1) = f(y).

This also shows injectivity by the above. f is surjective since if z ∈ [a, c),

z < b =⇒ f [g−1(z)/2] = z, z ≥ b =⇒ f[h−1(z) + 1]/2 = z.

Proof of (b). Suppose [x0, b) has order type [0, 1). For any i ∈ Z+, by (a), [xi, b) hasorder type [0, 1); applying (a) again gives that [xi, xi+1) has order type [0, 1).

Now suppose every [xi, xi+1) has order type [0, 1). If fi : [0, 1) → [xi, xi+1) areorder isomorphisms, first define

f : [0,∞)→ [x0, b), x 7→ fi(x− i) if x ∈ [i, i+ 1),

which is well-defined since any x ∈ [0,∞) is in some set of the form [i, i + 1). Weclaim f is an order isomorphism. If x, y ∈ [0,∞), then x ∈ [i, i+ 1), y ∈ [j, j + 1) forsome i, j. Suppose x < y. Then,

i 6= j =⇒ f(x) = fi(x− i) < xi+1 ≤ xj ≤ fj(y − j) = f(y),

i = j =⇒ f(x) = fi(x− i) < fi(y − i) = f(y),

since the fi are order-preserving; this also implies injectivity by the above. To showsurjectivity, we first know f maps onto

⋃i[xi, xi+1) by definition. So let z ∈ [x0, b).

Since b is the least upper bound of the xi, z is not an upper bound, so there exists isuch that z ∈ [xi, xi+1). But then, since the fi are bijective as well, f(f−1

i (z)+i) = z.Now let g : [0, 1) → [0,∞) be defined as x 7→ x/(1− x); this is an order isomor-

phism since it has inverse x/(1 + x), and since it is strictly monotonic increasing.Thus, f g : [0, 1)→ [x0, b) is a bijection, and preserves orientation since f, g do.

Proof of (c). Let a > a0. We proceed by transfinite induction. SΩ is a well-orderedset, and so if we let J be the set of a ∈ SΩ such that the claim holds, it suffices toshow that for every a ∈ J , Sa ⊆ J =⇒ a ∈ J .

24

We first show that either a has an immediate predecessor or there exists a se-quence ai ⊆ Sa such that a = supai. Suppose a does not have an immediatepredecessor. Then, we have the section Sa = bi, which is countable by definitionof SΩ. (b1, a] 6= ∅ since a has no immediate predecessor, and so let a1 ∈ (b1, a]. Weconstruct the ai inductively as follows: if we have an, let an+1 ∈ (supan, bn+1, a],which is nonempty as above. We then get a sequence of elements a1 < a2 < · · · < a.But since an > bn for all n by construction, we see that a ≥ supai. Moreover, ifa > supai, then supai = bk for some k, for Sa contains all elements less than a,and hence supai < ak, contradicting that supai is an upper bound.

Now suppose Sa ⊆ J . If a has an immediate predecessor a−1, then [a0×0, a×1) =[a0× 0, (a− 1)× 1)∪ [a× 0, a× 1) has order type [0, 1) by (a), for we have the orderisomorphism [a× 0, a× 1)→ [0, 1) defined by a× x 7→ x, which is trivially bijectiveand order-preserving since SΩ× [0, 1) was constructed with the dictionary order. Onthe other hand, if a does not have an immediate predecessor, then there exists asequence ai ⊆ Sa such that a = supai, and so the claim follows by (b).

Proof of (d). Let a× b, a′× b′ be two points in L; suppose without loss of generalitythat a×b < a′×b′. By (c), [a0×0, a×1) and [a0×0, a′×1) have order type [0, 1); by9a), this implies [a0×0, a×b) and [a0×0, a′×b′) have order type [0, 1). Hence, by (a),Y = [a× b, a′× b′) has order type [0, 1). Let f : [0, 1)→ Y be the order isomorphism.We claim f is continuous. First, since Y is an interval, it is convex, and so by Theorem16.4 the order topology on Y is the same as the subspace topology on Y inherited fromL. Then, for any basis set A = (c×d, c′×d′) ⊆ Y , f−1(A) = (f−1(c×d), f−1(c′×d′))since f is an order isomorphism, and moreover this preimage is open. Also, for anybasis set B = [a× b, c′× d′) ⊆ Y , f−1(B) = [f−1(a× b), f−1(c′× d′)), which is againopen. Thus, f is continuous. Finally, if we define

F (x) =

f(x) if x ∈ [0, 1)

a′ × b′ if x = 1

we have a continuous path F : [0, 1]→ [a× b, a′× b′] by the pasting lemma (Theorem18.3), and so L is path connected.

Proof of (e). Let a × b be a point in L. Since SΩ × [0, 1) does not have a maximalelement, there is some a′× b′ > a× b. Now by (c), there exists an order isomorphismf : [0, 1)→ [a0×0, a′× b′). Restricting f to (0, 1), we get another order isomorphismf : (0, 1) → (a0 × 0, a′ × b′). The set [a0 × 0, a′ × b′) is open in SΩ × [0, 1), and so(a0 × 0, a′ × b′) is open in L, and is a neighborhood of a× b.

We claim (a0 × 0, a′ × b′) is homeomorphic to (0, 1). We already have a bijectionthat is continuous by the same argument as in (d), and so it suffices to show f is

25

open as well. But if (x, y) ⊆ (0, 1) is a basis set, then f(x, y) = (f(x), f(y)) since fis an order isomorphism, and moreover open since the topology on (a0× 0, a′× b′) isthe order topology.

Proof of (f). Suppose L could be imbedded in Rn; then, every subspace of Rn has acountable basis by Example 30.1, and since L is homeomorphic with such a subspace,it also has a countable basis. Now, since X = (SΩ×0)\a0×0 is a convex subsetof L, the subspace topology on X is the same as the order topology by Theorem16.4. Thus, the intersection of the countable basis for L with X forms a countablebasis by Theorem 30.2. This implies that there is a countable subset Y of X that isdense in X by Theorem 30.3. By Theorem 10.3, though, this subset Y has an upperbound x in X. Thus, ∅ 6= (x,Ω) ⊆ X \ Y , and so the closure of Y is not all of X, acontradiction.

25 Components and Local Connectedness

Exercise 25.2.(a) What are the components and path components of Rω (in the product topology)?(b) Consider Rω in the uniform topology. Show that x and y lie in the same

component of Rω if and only if the sequence

x− y = (x1 − y1, x2 − y2, . . .)

is bounded.(c) Give Rω the box topology. Show that x and y lie in the same component of Rω

if and only if the sequence x− y is “eventually zero.”

Solution for (a). By Exercise 24.8(a), Rω is path connected, for Theorem 19.6 is notlimited to finite product topologies. Thus, Rω is the only path component, and soRω is the only component as well since path connected =⇒ connected.

Proof of (b). We first define ϕ : x 7→ x − y. We recall that since ρ(ϕ(x), ϕ(z)) =ρ(x, z), by Exercise 21.2 ϕ is an isometric imbedding that is moreover surjective (thepreimage of any z is z + y), ϕ is a homeomorphism. Thus, x − y is in the samecomponent as 0 if and only if x is in the same component as y, for ϕ, ϕ−1 do notmodify the topology of Rω.

It therefore suffices to check the case y = 0. Suppose x is bounded; then, wedefine f : [0, 1] → Rω where f(t) = (x1t, x2t, . . .). This is continuous since givenε > 0, B(f(t), ε) ⊇ f(B(t, ε/ sup|xn|)), where sup|xn| < ∞ by boundedness

26

of x. Thus, f connects 0 and x, i.e., they are in the same path component, andtherefore the same component by Theorem 25.5.

Conversely, recall by Exercise 23.8 that we have the separation Rω = A ∪ B,where A is the set of bounded sequences and B is the set of unbounded sequences ofreals. If x is unbounded it is in B and so is not in the same component as 0.

Proof of (c). x “eventually zero” here means that xi = 0 for all i ≥ N for some N .Note by the same argument as in (b), it suffices to consider the case y = 0.

Suppose first that x is not eventually zero. Define the function f = (fn), wherefn(a) = na/|xn| if xn 6= 0, and a otherwise. f is continuous since each fn is continuoussince it is linear, and so if f−1

n (Un) = Vn, we have f−1(∏Un) =

∏Vn. Note that

this is a bijection since each component has an inverse f−1n (a) = |xn|a/n if xn 6= 0,

and a otherwise, and moreover since the inverse is continuous since each componentis linear, we have a homeomorphism f : Rω → Rω. Since there are infinitely many nsuch that xn 6= 0, and so infinitely many n such that fn(xn) = n, we have that f(x) isunbounded, and thus, by the separation of Rω in the box topology in Example 23.6,we have that f(x) and 0 are in different components. Since f is a homeomorphism,this implies x and 0 are in different components as well.

Conversely, suppose x is eventually zero. Then, xn = 0 for all n ≥ N for someN , and so x ∈ RN × 0 × 0 × · · · ⊆ Rω; this subspace is homeomorphic toRN . Since RN is connected by Theorem 23.6, we see that x and 0 are in the samecomponent.

27 Compact Subspaces of the Real Line

Exercise 27.4. Show that a connected metric space having more than one point isuncountable.

Proof. Let X be a connected metric space with the metric d, and let x0, x1 ∈ Xbe distinct. Let d(x0, x1) = r, and define f(x) = d(x0, x). f is continuous by thediscussion on p. 175. We see that f(x0) = 0, f(x) = r, and so by the intermediatevalue theorem (Theorem 24.3), f(X) ⊇ [0, r], i.e., f maps onto [0, r].

Now suppose X is countable. Then, by Theorem 7.1 there exists a surjective func-tion g : Z+ → X, and so f g : Z+ → f(X) maps onto [0, r], which is a contradictionsince [0, r] is uncountable by Corollary 27.8.

29 Local Compactness

Exercise 29.4. Show that [0, 1]ω is not locally compact in the uniform topology.

27

Proof. Suppose X = [0, 1]ω is locally compact, and in particular at 0. Then, thereexists C compact that contains a neighborhood U 3 0. There exists X∩Bρ(0, ε) ⊆ U ;we see that 0, ε/3ω ⊆ X ∩Bρ(0, ε). 0, ε/3ω is closed since

0, ε/3ω =∏0, ε/3 =

∏0, ε/3 =

∏0, ε/3 = 0, ε/3ω

in the product topology by Theorem 19.5, which is finer than the uniform topology,and so it is compact by Theorem 26.2 since it is a closed subset of C compact, i.e.,limit point compact by Theorem 28.2.

We claim this is a contradiction. Consider x ∈ X, and the ball X ∩ Bρ(x, ε/9).Note that the distance between any two distinct points of 0, ε/3ω is ε/3, and sosince the diameter of X ∩Bρ(x, ε/9) is at most 2ε/9, X ∩Bρ(x, ε/9) contains at mostone point of 0, ε/3ω. Thus, 0, ε/3ω contains no limit points, and so is not limitpoint compact, a contradiction.

Exercise 29.8. Show that the one-point compactification of Z+ is homeomorphic tothe subspace 0 ∪ 1/n | n ∈ Z+ of R.

Proof. Let K = 1/n | n ∈ Z+. Let f : R+ → R+ such that f(x) = 1/x; this is ahomeomorphism since it is continuous and is its own inverse. By Theorem 18.2(d)and 18.2(e), f : Z+ → f(Z+) = K is continuous, and again is a homeomorphismsince it is its own inverse. Now consider Y = 0 ∪K, which is closed and boundedand therefore compact by Theorem 27.3, and Hausdorff by Theorem 17.11. SinceK ′ = 0 by Example 17.8, we know Y is the one-point compactification of K. IfX = p ∪ Z+ is the one-point compactification of Z+, and letting g : p 7→ 0 ∈ Y ,which is clearly continuous, the function h : X → Y defined by the pasting lemma(Theorem 18.3) applied to f, g is also continuous, and has continuous inverse definedby the pasting lemma applied to f−1, g−1, and so is a homeomorphism X ↔ Y .

4 Countability and Separation Axioms

30 The Countability Axioms

Exercise 30.4. Every compact metrizable space X has a countable basis.

Proof. For given n ∈ Z+, we have an open cover of X by B(x, 1/n) for each x ∈ X;since X is compact, let An be the finite subcover.

⋃nAn is countable since it is the

countable union of finite sets; we claim it is a basis for X. Let U ⊆ X open andx ∈ U . Since X is metrizable, there exists B(x, δ) ⊆ U for some δ > 0. Let N such

28

that 2/N < δ. Since AN covers X, there exists B(y, 1/N) 3 x. B(y, 1/N) ⊆ B(x, δ),for if we choose z ∈ B(y, 1/N), d(x, z) ≤ d(x, y) + d(y, z) ≤ 1/N + 1/N < δ. Thus,x ∈ B(y, 1/N) ⊆ U , and so

⋃nAn is a countable basis by Lemma 13.2.

Exercise 30.5.(a) Show that every metrizable space with a countable dense subset has a countable

basis.(b) Show that every metrizable Lindelof space has a countable basis.

Proof of (a). Let X be a metrizable space and A a countable dense subset. We claimthat the set of open balls in X below is a basis for X:

B := B(a, 1/n) ⊆ X | a ∈ A, n ∈ N.

Note B is countable since is in bijection with A×N. So let x be contained in an opensubset U ⊆ X; since X is metrizable, x ∈ B(x, ε) ⊆ U for some small ε. Let n besuch that 1/n < ε/2. Then, since A is dense, some a ∈ A is contained in B(x, 1/n),and conversely x ∈ B(a, 1/n). By the triangle inequality, x ∈ B(a, 1/n) ⊆ U , so byLemma 13.2 we are done.

Proof of (b). Let X be a metrizable space. Then, the set of open balls

Bn := B(x, 1/n) ⊆ X | x ∈ X

is an open cover of X for each n ∈ N; since X is Lindelof, it has a countablesubcover Bn. We claim B :=

⋃n∈N Bn is a basis for X; note it is countable since it is

a countable union of countable sets. So let x ∈ B(x, ε) ⊆ U as before, and let n suchthat 1/n < ε/2. Then, there is some x′ ∈ X such that B(x′, 1/n) ∈ Bn contains x.By the triangle inequality, x ∈ B(x′, 1/n) ⊆ U , so by Lemma 13.2 we are done.

Exercise 30.8. Which of our four countability axioms does Rω in the uniform topol-ogy satisfy?

Solution. Rω is first countable since it is metrizable (see p. 130 and Example 30.2),but is not second countable by Example 30.2. By Exercise 30.5, we then see that Rω

does not have a countable dense subset, and is also not Lindelof.

Exercise 30.9. Let A be a closed subspace of X. Show that if X is Lindelof, thenA is Lindelof. Show by example that if X has a countable dense subset, A need nothave a countable dense subset.

29

Proof. X is Lindelof if and only if a collection of closed subsets of X with emptyintersection has a countable subcollection with empty intersection by taking comple-ments in Theorem 30.3(a). Now suppose we have a collection C of closed subsets of Awith empty intersection; it is then also a collection of closed subsets of X with emptyintersection by Theorem 17.3 since A is closed, and so has a countable subcollectionwith empty intersection since X is Lindelof. Thus, A is also Lindelof.

Now let X = R2` . We see Q2 is countable, and is dense in X since if we take

x ∈ X and a neighborhood U 3 x, there exists a basis element [a, b) × [c, d) ⊆ Ucontaining x, which intersects Q2 by the fact that (a, b)× (c, d) ∩Q2 6= ∅ since Q isdense in R. Thus, X has a countable dense subset; we claim that L = x × (−x) |x ∈ R` is a closed subspace of X that does not have a countable dense subset.L is closed since if (x1, x2) ∈ X \ L, then letting d = x1 + x2, the basis element[x1 − d/3, x1 + d/3), [x2 − d/3, x2 + d/3) does not intersect L. But then, L has thediscrete topology since (x,−x) = L∩ [x, b)× [−x, d) is open in L. Thus, if A ⊆ L,A = A by discreteness, and so A = L is true if and only if A = L. Thus, L has nocountable dense subset.

Exercise 30.17. Give Rω the box topology. Let Q∞ denote the subspace consistingof sequences of rationals that end in an infinite string of 0’s. Which of our fourcountability axioms does this space satisfy?

Proof. We claim Q∞ is not first countable, and therefore not second countable. Sup-pose we have a countable basis Ui at 0 = (0, 0, 0, . . .) ∈ Qω. Let Vj ( πj(Uj) openin Q with the subspace topology induced by R. Then, the neighborhood

∏j Vj 3 0

does not contain any Ui, so Ui is not a basis and Q∞ is not first or second countable.We now show Q∞ has a countable dense subset. For, Qn = Qn × 0 × 0 are

countable since they are finite products of countable sets, and so their countableunion Q∞ =

⋃Qn is also countable. Thus, Q∞ is countable and so is a countable

dense subset of itself.We now show Q∞ is Lindelof. Suppose V is an open covering of Q∞. Then, since

Q∞ is countable, choosing for every x ∈ Q∞ one element V ∈ V such that x ∈ V ,we get a countable subcover of Q∞.

31 The Separation Axioms

Exercise 31.3. Show that every order topology is regular.

Proof. Let X be an ordered set with the order topology. X is Hausdorff and thereforeT1 by Theorem 17.11. It therefore suffices to show the condition in Lemma 31.1(a).

30

So let x ∈ X and let U be an open neighborhood of x; we will construct a basiselement V of the order topology such that x ∈ V and V ⊆ U .

Suppose x is neither the smallest nor largest element of X. Then, x ∈ (a, b) ⊆ Ufor some basis element (a, b) of the order topology. If (a, x), (x, b) are nonempty,then let V = (u, v) where u ∈ (a, x) and v ∈ (x, b). If (a, x) = ∅, let u = a, sothat V = (u, v) = [x, v); if (x, b) = ∅, let v = b, so that V = (u, v) = (u, x]. Then,x ∈ (u, v) and (u, v) ⊆ (a, b) ⊆ U .

Now suppose x is the smallest (resp. largest) element of X. Then, x ∈ [x, b) ⊆ U(resp. x ∈ (a, x] ⊆ U) for some basis element [x, b) (resp. (a, x]) of the order topology.Now if (x, b) (resp. (a, x)) is nonempty, then let V = [x, v) (resp. (u, x]) wherev ∈ (x, b) (resp. u ∈ (a, x)). On the other hand, if (x, b) (resp. (a, x)) is empty, thenlet v = b (resp. u = a), so that V = x. We then have x ∈ V and V ⊆ U .

If x is the smallest and largest element of X, then X = x is trivially regular.

32 Normal Spaces

Exercise 32.1. Show that a closed subspace of a normal space is normal.

Proof. Suppose Y is our closed subspace of our normal space X, and A,B ⊆ Ydisjoint and closed. By Theorem 17.3, A,B are closed in X. Let U, V be a separationof A,B in X. Then, Y ∩ U, Y ∩ V separate A,B in Y , and so Y is normal.

Exercise 32.3. Show that every locally compact Hausdorff space is regular.

Proof. Let X be a locally compact Hausdorff space; in particular it is T1. Let x ∈ Xwith neighborhood U 3 x. By Theorem 29.2, there exists a neighborhood V 3 xsuch that V ⊆ U . By Lemma 31.1(a), X is then regular.

Exercise 32.4. Show that every regular Lindelof space is normal.

Proof. Let A,B be disjoint closed subsets of X regular and Lindelof. For all x ∈ A,there exists a neighborhood U 3 x disjoint from B. By regularity, there exists aneighborhood U ⊆ U ⊆ U containing x; since these U cover A, and A is Lindelof byExercise §30.9, there exists a countable subcover Ui such that Ui ∩B = ∅ for all i.Similarly, we can construct a countable subcover Vi of B such that Vi ∩A = ∅ forall i. By the exact same argument as in the proof of Theorem 32.1, then, the sets

U ′ =⋃n∈Z+

(Un \

n⋃i=1

Vi

), V ′ =

⋃n∈Z+

(Vn \

n⋃i=1

Ui

)are open and U ′ ⊇ A, V ′ ⊇ B,U ′ ∩ V ′ = ∅, and so X is normal.

31

Exercise 32.5. Is Rω normal in the product topology? In the uniform topology?

Proof. Since Rω is metrizable in the product topology by Theorem 20.5 and in theuniform topology by definition on p. 124, both are normal by Theorem 32.2.

33 The Urysohn Lemma

Exercise 33.1. Examine the proof of the Urysohn lemma, and show that for givenr,

f−1(r) =⋂p>r

Up −⋃q<r

Uq,

p, q rational.

Proof. ⊆. Suppose x ∈ f−1(r), i.e., x ∈ Ur by definition, and x /∈ Uq for all q < r bydefinition in Step 3. Then, x ∈ Ur ⊆ Up for all p > r by construction in Steps 1, 2.⊇. Suppose x ∈

⋂p>r Up −

⋃q<r Uq. This implies x ∈ Up ⊆ Up for all p > r, and

so f(x) ≤ r by Step 4(1), and also x /∈ Uq for all q < r, and so f(x) ≥ r by Step4(2). Thus, f(x) = r.

34 The Urysohn Metrization Theorem

Exercise 34.3. Let X be a compact Hausdorff space. Show that X is metrizable ifand only if X has a countable basis.

Proof. ⇒. X is compact and metrizable, hence second countable by Exercise 30.4.⇐. X is compact and Hausdorff, and so X is regular by Theorem 32.3. X

is second countable as well, and so by the Urysohn metrization theorem (Theorem34.1), X is metrizable.

Exercise 34.5. Let X be a locally compact Hausdorff space. Let Y be the one-pointcompactification of X. Is it true that if X has a countable basis, then Y is metrizable?Is it true that if Y is metrizable, then X has a countable basis?

Solution. If Y is metrizable, then it is second countable by Exercise 30.4. X is thensecond countable by Theorem 30.2.

Now suppose X has a countable basis B = Bi. We claim that B with sets ofthe form Y \

⋃Bi, where the union is finite and the closure is taken in Y , form a

basis of Y ; call this larger basis B+. The Bi ∈ B are open by Lemma 16.2, andthe Y \

⋃Bi by Theorem 26.3. B+ is countable since B is countable and there are

countably many Y \⋃Bi by Exercise 7.5(j).

32

Now recall that we have two types of open sets of Y , sets U open in X, andsets Y \ C for C compact in X, by construction in Theorem 29.1. By Lemma 13.2,it suffices to show that for each x in an open set, we can find an element of B+

containing x properly contained in the open set. So first consider the U . For everyx ∈ U , there exists Bi ∈ B such that x ∈ Bi ⊆ U since B is a basis of X. Nowconsider the Y \ C and x ∈ Y \ C. If x ∈ X, then Y \ C ∩ X 3 x is open in Xby Theorem 29.1 (which says X is a subspace), and so the previous argument forU ⊆ X applies. It remains to show the case x =∞, where ∞ = Y \X. For eachy ∈ C, we can find a neighborhood V 3 y in X such that V ⊆ X = Y \ x bylocal compactness. Since V is open in X, there exists some Bi ⊆ V containing y.These Bi cover C and so there is a finite subcover by compactness C ⊆

⋃Bi; also,

x /∈⋃Bi ⊇ C. Thus, we have x ∈ Y \

⋃Bi ⊆ Y \ C. B+ is therefore a basis for Y

by Lemma 13.2.Since Y is compact and Hausdorff, it is normal by Theorem 32.3, and in particular,

Y is regular. Finally, since Y also has a countable basis, it is metrizable by theUrysohn metrization theorem (Theorem 34.1).

36 Imbeddings of Manifolds

Exercise 36.1. Prove that every manifold is regular and hence metrizable. Wheredo you use the Hausdorff condition?

Proof. Let X be our m-manifold. We first show X is locally compact. Let x ∈ X anda neighborhood U 3 x be given. SinceX is a manifold, there exists a homeomorphismf : U → f(U) ⊆ Rm. Since Rm is locally compact by Example 29.2, there exists aneighborhood V ⊆ f(U) of f(x) such that V is compact and V ⊆ f(U) by Theorem29.2. Then, x ∈ f−1(V ) ⊆ f−1(V ) ⊆ U . But then, f−1(V ) is compact and thereforeclosed by Theorems 26.3 and 26.5, and so f−1(V ) ⊆ f−1(V ) since the closure of a setis the intersection of all closed sets containing it, and moreover f−1(V ) = f−1(V ) byTheorem 18.1. Finally, we have x ∈ f−1(V ) ⊆ f−1(V ) ⊆ U , with f−1(V ) compact,and so X is locally compact by Theorem 29.2.

Now since X is locally compact and Hausdorff, X is regular by Exercise 32.3.Since X is regular and has a countable basis, it is metrizable by the Urysohn metriza-tion theorem (Theorem 34.1). Note we used that X is Hausdorff in showing X isregular, for the characterization of local compactness, and the assertion that compactimplies closed.

Exercise 36.5. The Hausdorff condition is an essential part of the definition of amanifold; it is not implied by the other parts of the definition. Consider the following

33

space: Let X be the union of the set R−0 and the two-point set p, q. TopologizeX by taking as basis the collection of all open intervals in R that do not contain0, along with all sets of the form (−a, 0) ∪ p ∪ (0, a) and all sets of the form(−a, 0) ∪ q ∪ (0, a), for a > 0. The space X is called the line with two origins.

(a) Check that this is a basis for a topology.(b) Show that each of the spaces X − p and X − q is homeomorphic to R.(c) Show that X satisfies the T1 axiom, but is not Hausdorff.(d) Show that X satisfies all the conditions for a 1-manifold except for the Haus-

dorff condition.

Proof of (a). For any x ∈ X, for a large enough either (−a, 0) ∪ p ∪ (0, a) or(−a, 0) ∪ q ∪ (0, a) contains x. Moreover, if we have two basis elements B1, B2,their intersection is either empty, already another basis element, or a set of the form(−a, 0)∪ (0, a), we have a basis for X, for if x ∈ B1 ∩B2, B1 ∩B2, or in the last casechoosing (−a, 0) or (0, a), would be a basis element containing x that is contained inthe intersection.

Proof of (b). Let f : X \ q → R be defined such that x 7→ 0 if x = p, and x 7→ xotherwise. Clearly f is a bijection; it suffices to show it is continuous and open.A basis element not containing p, q maps to a basis element not containing 0 bydefinition and vice versa. A basis element of the form (−a, 0) ∪ p ∪ (0, a) mapsto (−a, a), and in the other direction, an open interval (−a, b) for a, b > 0 maps to(−a, 0)∪ p ∪ (0, a)∪ (0, b) if a < b, and similarly if a > b. Thus, f is a homeomor-phism. Note g : X \ p → R is also a homemorphism by the same argument.

Proof of (c). To show X is T1, it suffices to show x is closed for all x ∈ X. p, qare closed since X \ p = (−∞, 0)∪q∪ (0,∞), X \ q = (−∞, 0)∪p∪ (0,∞)are open. x is closed for x 6= p, q since if x < 0, X \ x = (−∞, x) ∪ (x, 0) ∪p, q ∪ (0,∞) is open, and likewise for if x > 0. Note X is not Hausdorff since anyneighborhood U of p intersects any neighborhood V of q, since U, V must containbasis elements Bp, Bq that contain p, q respectively; however, Bp ∩ Bq = ((−a, 0) ∪p ∪ (0, a)) ∩ ((−b, 0) ∪ p ∪ (0, b)) = (−c, 0) ∪ (0, c) 6= ∅ for c = mina, b.

Proof of (d). We claim the basis elements with rational end points form a countablebasis. For any (a, b) basis element containing x, we can find a < r < s < b such thatx ∈ (r, s) ⊆ (a, b); for any (−a, 0) ∪ p ∪ (0, a) basis element containing x, we canfind 0 < r < a such that x ∈ (−r, 0)∪p∪ (0, r) ⊆ (−a, 0)∪p∪ (0, a). Thus theyform a basis by Lemma 13.2.

Now for any x 6= p, q, we see that there is a neighborhood U of x not contain-ing p, q, and U is homeomorphic to a neighborhood in R by (b). For x = p, q,

34

any neighborhood of x will contain a basis element (−a, 0) ∪ x ∪ (0, a), which ishomeomorphic to (−a, a) by (b).

Part II

Algebraic Topology

9 The Fundamental Group

51 Homotopy of Paths

Exercise 51.1. Show that if h, h′ : X → Y are homotopic and k, k′ : Y → Z arehomotopic, then k h and k′ h′ are homotopic.

Proof. Let F be the homotopy between h, h′, and G the homotopy between k, k′.Let H : X × 1 → Z where H(x, t) = G(F (x, t), t). Then H(x, 0) = G(F (x, 0), 0) =G(f(x), 0) = (k h)(x), H(x, 1) = G(F (x, 1), 1) = G(f(x), 1) = (k′ h′)(x).

It remains to show H is continuous. H is the map (x, t) 7→ (F (x, t), t) 7→G(F (x, t), t); since G is already continuous and the composition of continuous func-tions is continuous, it suffices to show (x, t) 7→ (F (x, t), t) is continuous. But this isclear since this map is continuous in each coordinate in the codomain.

Exercise 51.2. Given spaces X and Y , let [X, Y ] denote the set of homotopy classesof maps of X into Y .

(a) Let I = [0, 1]. Show that for any X, the set [X, I] has a single element.(b) Show that if Y is path connected, the set [I, Y ] has a single element.

Proof of (a). Fix f0 : X → I. For arbitrary f : X → I, f ' f0 by straight-linehomotopy since any straight line in I is contained in I. Thus, since f was arbitrary,[X, I] = [f0].

Proof of (b). Let f : I → Y and y = f(0). Let g : I → Y be the constant map withvalue y ∈ Y . Define F : I × I → Y by F (s, t) = f(s(1 − t)). Since F (s, 0) = f(s)and F (s, 1) = g(s), we see F is a homotopy f ' g. Now fix y0 ∈ Y , and let ρ be apath connecting y, y0. Define H : X × I → Y such that H(s, t) = ρ(t). Then, if f0

is the constant map with value y0 ∈ Y , H is a homotopy g ' f0. By transitivity ofhomotopy, we see f ' f0, and so since f was arbitrary, [X, Y ] = [f0].

Exercise 51.3. A space X is said to be contractible if the identity map iX : X → Xis nulhomotopic.

35

(a) Show that I and R are contractible.(b) Show that a contractible space is path connected.(c) Show that if Y is contractible, then for any X, the set [X, Y ] has a single

element.(d) Show that if X is contractible and Y is path connected, then [X, Y ] has a single

element.

Proof of (a). Since both I and R are convex, we see that any two maps are homotopicby straight-line homotopy as on pp. 324–325 since then the straight lines are fullycontained in convex set. In particular, iX ' f for any constant map f .

Proof of (b). If X is our contractible space, and F is our homotopy between iX andf0 for f0 a constant map with value x0 ∈ X, the map ρ : I → X where ρ(t) = F (x, t)is a path connecting x, x0 for any x ∈ X.

Proof of (c). Let g0 : Y → Y be the constant map with value y0 ∈ Y ; we haveiY ' g0. Now define f0 : X → Y such that f0(x) = y0. Then, for any f : X → Y ,we have f = iY f ' g0 f = f0 by Exercise 51.1, and so since f was arbitrary,[X, Y ] = [f0].

Proof of (d). Let g0 : X → X be the constant map with value x0 ∈ X; we haveiX ' g0. Now define y = f(x0). Then, for any f : X → Y , we have f = fiX ' fg0,which is the constant map at y. Fix y0 ∈ Y with f0 : X → Y the constant map withvalue y0 ∈ Y , and let ρ be a path connecting y, y0. Define H : X × I → Y such thatH(x, t) = ρ(t). Then, H is a homotopy between f g0 and f0. By transitivity ofhomotopy, we see f ' f0, and so since f was arbitrary, [X, Y ] = [f0].

52 The Fundamental Group

Exercise 52.4. Let A ⊂ X; suppose r : X → A is a continuous map such thatr(a) = a for each a ∈ A. (The map r is called a retraction of X onto A.) Ifa0 ∈ A, show that

r∗ : π1(X, a0) −→ π1(A, a0)

is surjective.

Proof. Letting ι : A → X be the inclusion map, we see rι = idA, and so by Theorem52.4, r∗ ι∗ = (r ι)∗ = (idA)∗ = idπ1(A,a0). This implies r∗ is surjective.

36

53 Covering Spaces

Exercise 53.3. Let p : E → B be a covering map; let B be connected. Show that ifp−1(b0) has k elements for some b0 ∈ B, then p−1(b) has k elements for every b ∈ B.In such a case, E is called a k-fold covering of B.

Proof. Since p is a covering map, for any b with |p−1(b)| = j there exists a neigh-borhood U 3 b such that p−1(U) is the disjoint union of j open neighborhoods Vαhomeomorphic to U , for each Vα contains a unique preimage of b, which must be onepoint since U, Vα are homeomorphic. By the same argument, all x ∈ U are such that|p−1(x)| = j. Thus, we can partition B into disjoint open sets Aj where each x ∈ Ajare such that p−1(x) = j (note that j can be any cardinal since the argument abovedoes not depend on j being finite), and Ak 3 b0 is one of the Aj; however, since B isconnected, B = Ak, for otherwise the Aj will be a separation of B.

Exercise 53.5. Show that the map of Example 3 is a covering map. Generalize tothe map p(z) = zn.

Proof. Note that p(z) when considered as a map on S1 as a subset of R2 is given byp(cos 2πx, sin 2πx) = (cos 2πnx, sin 2πnx). Letting q(x) = (2 cosπx, sin 2πx) be thecovering R → S1 in Theorem 53.1, and r : R → R be the multiplication by n, wethen have the commutative diagram

R2 R2

S1 S1

r

q

p

q−1

Now let x ∈ S1, and consider U = S1 \ x. Taking s ∈ q−1(x), we see thatwe have a partition q−1(U) =

⊔i∈ZWi where Wi = (s + i, s + i + 1), and we also

have r−1(Wi) = ((s + i)/n, (s + i + 1)/n) =: Zi. Letting Vi = q(Zi), we see thatp−1(U) =

⊔0≤i<n Vi. Thus, for all i,

Wi Zi

Vi U

r|Wi

q|Zi

p|Vi

(q|Wi)−1

commutes, where the top and vertical arrows are homeomorphisms, and so we havea homeomorphism between Vi and U for all i.

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54 The Fundamental Group of the Circle

Exercise 54.1. What goes wrong with the “path-lifting lemma” (Lemma 54.1) forthe local homeomorphism of Example 2 of §53?

Solution. Since b0 = (1, 0) cannot be covered by an open set that is evenly coveredby p by Example 53.2, we see that the first step in the proof of Lemma 54.1 fails.

Exercise 54.4. Consider the covering map p : R × R+ → R2 − 0 of Example 6 of§53. Find liftings of the paths

f(t) = (2− t, 0),

g(t) = ((1 + t) cos 2πt, (1 + t) sin 2πt),

h(t) = f ∗ g.

Sketch these paths and their liftings.

Solution. We see first that the covering map p is the mapping

(x, s) 7→ ((cos 2πx, sin 2πx), s) 7→ s(cos 2πx, sin 2πx).

Thus, we have the family of liftings

fn(t) = (n, 2− t)gn(t) = (t+ n, 1 + t)

hn(t) =

(n, 2− 2t) if t ∈ [0, 1/2]

(2t+ n− 1, 2t) if t ∈ [1/2, 1]

where n ∈ Z. We omit the sketches.

Exercise 54.5. Consider the covering map p × p : R × R → S1 × S1 of Example 4of §53. Consider the path

f(t) = (cos 2πt, sin 2πt)× (cos 4πt, sin 4πt)

in S1 × S1. Sketch what f looks like when S1 × S1 is identified with the doughnutsurface D. Find a lifting f of f to R× R, and sketch it.

Solution. We have the family of liftings fn(t) = (n + t/2, n + t), where n ∈ Z. Weomit the sketches.

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58 Deformation Retracts and Homotopy Type

Exercise 58.1. Show that if A is a deformation retract of X, and B is a deformationretract of A, then B is a deformation retract of X.

Proof. Let F : X × I → X, G : A× I → A be the deformation retractions of X ontoA and A onto B respectively. We claim that

H(x, t) =

F (x, 2t) if 0 ≤ t ≤ 1/2

ι G(F (x, 1), 2t− 1) if 1/2 ≤ t ≤ 1

is a deformation retraction of X onto B, where ι : A → X is the inclusion map.We see H(x, 0) = F (x, 0) = x, and that if x ∈ B, H(x, 1) = ι G(F (x, 1), 1) =ι F (x, 1) = ι x = x; moreover, H is continuous by the pasting lemma, sinceH(x, 1/2) = F (x, 1) = ι G(F (x, 1), 0) = ι F (x, 1) = F (x, 1), and since ι G(F (x, 1), 2t− 1) is a composition of continuous functions.

Exercise 58.2. For each of the following spaces, the fundamental group is eithertrivial, infinite cyclic, or isomorphic to the fundamental group of the figure-eight.Determine for each space which of the three alternatives holds.

(a) The “solid torus,” B2 × S1.(b) The torus T with a point removed.(c) The cylinder S1 × R.(d) The infinite cylinder S1 × R.(e) R3 with the nonnegative x, y, and z axes deleted.

The following subsets of R2:(f) x | ‖x‖ > 1(g) x | ‖x‖ ≥ 1(h) x | ‖x‖ < 1(i) S1 ∪ (R+ × 0)(j) S1 ∪ (R+ × R)(k) S1 ∪ (R× 0)(l) R2 − (R+ × 0)

Remark. We use Theorem 60.1 to say that π1(−) and × commute.

Solution. (a). π1(B2 × S1) = π1(B2) × π1(S1) = π1(S1), for we have a deformationretraction from B2 onto 0.

(b). π1(T \ p) is isomorphic to the fundamental group of the figure-eight, forwe have a deformation retraction onto the figure-eight by deforming T \ p in thefollowing manner:

39

a a

b

b

a a

b

b

a a

b

b

a b

where the last arrow comes from the construction of the torus as the quotient R2/Z2.(c). π1(S1×I) = π1(S1)×π1(I) = π1(S1), for we have the deformation retraction

from I onto 0.(d). π1(S1 × R) = π1(S1) × π1(R), for we have the deformation retraction from

R onto 0.(e). This space retracts onto S2 \ p, q, r ' R2 \ s, t (where the isomorphism

is from Theorem 59.3), which retracts onto the figure-eight, and so the fundamentalgroup is isomorphic to the fundamental group of the figure-eight.

(f). π1(x | ‖x‖ > 1) = π1(S1), by a deformation retraction onto a circle ofradius > 1, whose fundamental group is π1(S1) since it is homeomorphic to S1.

(g). π1(x | ‖x‖ ≥ 1) = πs(S1), for we have the deformation retraction onto S1.

(h). π1(x | ‖x‖ < 1) = π1(0) = 1, by a deformation retraction onto 0.(i). π1(S1∪ (R+× 0)) = π1(S1), since we can retract R+ to one point (1, 0) ∈ S1.(j). π1(S1 ∪ (R+ × R)) = π1(S1), since we can retract R+ × R to the half-circle

S1 ∩ R+ × R.(k). π1(S1 ∪ (R× 0)) is isomorphic to the fundamental group of the figure-eight,

for we can retract (R × 0) onto the line segment from (−1, 0) to (1, 0), which givesa topological space homotopy equivalent to the figure-eight.

(l). π1(R2 \ (R+ × 0)) = 1, since we can retract onto any arbitrary point p ∈R2 \ (R+ × 0).

Exercise 58.9. We define the degree of a continuous map h : S1 → S1 as follows:Let b0 be the point (1, 0) of S1; choose a generator γ for the infinite cyclic group

π1(S1, b0). If x0 is any point of S1, choose a path α in S1 from b0 to x0, and defineγ(x0) = α(γ). Then γ(x0) generates π(S1, x0). The element γ(x0) is independent ofthe choice of the path α, since the fundamental group of S1 is abelian.

Now given h : S1 → S1, choose x0 ∈ S1 and let h(x0) = x1. Consider thehomomorphism

h∗ : π(S1, x0) −→ π1(S1, x1).

Since both groups are infinite cyclic, we have

h∗(γ(x0)) = d · γ(x1) (∗)

40

for some integer d, if the group is written additively. The integer d is called thedegree of h and is denoted by deg h.

The degree of h is independent of the choice of the generator γ; choosing the othergenerator would merely change the sign of both sides of (∗).

(a) Show that d is independent of the choice of x0.(b) Show that if h, k : S1 → S1 are homotopic, they have the same degree.(c) Show that deg(h k) = (deg h) · (deg k).(d) Compute the degrees of the constant map, the identity map, the reflection map

ρ(x1, x2) = (x1,−x2), and the map h(z) = zn, where z is a complex number.(e) Show that if h, k : S1 → S1 have the same degree, they are homotopic.

Proof of (a). Let x0 6= y0 ∈ S1, and let h(y0) = y1. Let α be a path from x0 to y0;

then, β′ = h β is a path from x1 to y1. Then, γ(y1) = [β′] ∗ γ(x1) ∗ [β′], and so

h∗(γ(y0)) = h∗([β] ∗ γ(x0) ∗ [β]) = [β′] ∗ h∗(γ(x0)) ∗ [β′]

= [β′] ∗ d · γ(x1) ∗ [β′] = d · ([β′] ∗ γ(x1) ∗ [β′]) = d · γ(y1).

Proof of (b). Choose x0 ∈ S1 and let h(x0) = y0, k(x0) = y1. By Lemma 58.4, thereexists a path α from y0 to y1 such that k∗ = α h∗. Then, we have

deg k · γ(y0) = k∗(γ(x0)) = α(h∗(γ(x0)))

= α(deg h · γ(y1)) = deg h · α(γ(y1)) = deg h · γ(y0),

and so deg k = deg h.

Proof of (c). Choose x0 ∈ S1 and let k(x0) = y0, h(y0) = y1. Then,

deg(h k) · γ(y1) = (h k)∗(γ(x0)) = (h∗ k∗)(γ(x0))

= h∗(deg k · γ(y0)) = deg k · h∗(γ(y0)) = (deg h) · (deg k)γ(y1),

and so deg(h k) = (deg h) · (deg k).

Solution for (d). Let k : S1 → S1 be the constant map. Then, we have k∗(γ(x0)) = 0,and so deg k = 0.

Let id : S1 → S1 be the constant map. Then, we have id∗(γ(x0)) = γ(x0), and sodeg id = 1.

Let h : S1 → S1 such that z 7→ zn. Let p : R → S1 be the standard covering.p−1(b0) = 0, which is fixed by h. Let γ be the loop γ(s) = e2πis, which generatesπ1(S1, b0); note we can choose γ, b0 in this way by the fact that degree is independentof choice of generator and x0. Then, (h γ)(s) = e2πins = p(ns), and so the lift ofh γ is s 7→ ns, i.e., h∗(γ(b0)) = nγ(b0). Thus, deg h = n.

The case for ρ follows from that of h, where n = −1.

41

Proof of (e). By (a), we use b0 as our base point, and let h(b0) = x0, k(b0) = y0.Consider h∗ : π1(S1, b0) → π1(S1, x0), k∗ : π1(S1, b0) → π1(S1, y0); by assumption,there exists n, such that h∗(γ(b0)) = n · γ(x0), k∗(γ(b0)) = n · γ(y0).

Now let γ(s) = e2πis, which generates π1(S1, b0) as above. Then, let h γ, k γ bethe lifts of hγ, kγ through the standard covering p : R→ S1, at h0 ∈ p−1(x0), k0 ∈p−1(y0) respectively. Let h1 = h γ(1), k1 = k γ(1).

Now we have h∗(γ(b0)) = (h1 − h0)γ(x0), k∗(γ(b0)) = (k1 − k0)γ(y0), and so

h1 − h0 = k1 − k0 = n. We want to find a path homotopy from h γ to k γ. We

first define˜h γ = h γ−h0 +k0 so that both

˜h γ, h γ have the same end points;

we then see there exists a path homotopy F between them since R is contractible.

Now consider h(x) = e2πi(k0−h0)h(x). Then,˜h γ is the lift of h γ at k0. Then,

p F is a path homotopy from˜h γ to k γ. Since this is a path homotopy, it factors

through (p, id) to get a homotopy F from h to k.We now see that there exists a path homotopy G between h and h given by

G(x, t) = h(x)e2πis(k0−h0), and so there exists a path homotopy combining F,G be-tween h and k.

Exercise 58.10. Suppose that to every map h : Sn → Sn we have assigned an integer,denoted by deg h and called the degree of h, such that

(i) Homotopic maps have the same degree.(ii) deg(h k) = (deg h) · (deg k).

(iii) The identity has degree 1, any constant map has degree 0, and the reflectionmap ρ(x1, . . . , xn+1) = (x1, . . . , xn,−xn+1) has degree −1.

Prove the following:(a) There is no retraction r : Bn+1 → Sn.(b) If h : Sn → Sn has degree different from (−1)n+1, then h has a fixed point.(c) If h : Sn → Sn has degree different from 1, then h maps some point x to is

antipode −x.(d) If Sn has a nonvanishing tangent vector field v, then n is odd.

Proof of (a). We use polar coordinates to denote Bn+1 and Sn, i.e., the points inBn+1 are given by (θ, φ2, . . . , φn, ρ) and the points in Sn by (θ, φ2, . . . , φn), where0 ≤ θ < 2π, 0 ≤ φi < π, and 0 ≤ ρ ≤ 1. Thus, the inclusion Sn → Bn+1 is given by(θ, φ2, . . . , φn) 7→ (θ, φ2, . . . , φn, 1).

Now, suppose a retraction r : Bn+1 → Sn exists. Then, define the homotopyH : Sn × [0, 1]→ Sn by H(x1, . . . , xn, t) = r(x1, . . . , xn, t), which is continuous sincer is. Then, H(x1, . . . , xn, 0) is constant, hence has degree 0, whereas H(x1, . . . , xn, 1)is the identity, hence has degree 1, which contradicts (i).

42

Proof of (b). We return to Cartesian coordinates. Suppose h has no fixed points.We first construct a homotopy:

H(x, t) =(1− t)h(x) + ta(x)

‖(1− t)h(x) + ta(x)‖.

Since H is continuous since it is a composition of continuous functions in bothvariables, it suffices to show ‖(1 − t)h(x) + ta(x)‖ 6= 0 for all x, t. So, suppose(1 − t)h(x) + ta(x) = 0. Then, (1 − t)h(x) = tx. Comparing norms, this is onlypossible when t = 1/2, and so h(x) = x, which contradicts that h has no fixed points.

Thus, we see that h is homotopic to the antipodal map. Now, since the antipodalmap is the composition of n+1 reflections, each one reflecting each coordinate of Sn,we have that deg(h) = deg(a) = (−1)n+1 by (i), (ii), and (iii), a contradiction.

Proof of (c). Suppose h maps no point x to its antipode −x. We first construct ahomotopy:

G(x, t) =(1− t)h(x) + tx

‖(1− t)h(x) + tx‖.

Since G is continuous since it is a composition of continuous functions in bothvariables, it suffices to show ‖(1 − t)h(x) + tx‖ 6= 0 for all x, t. So, suppose(1 − t)h(x) + tx = 0. Then, (1 − t)h(x) = −tx. Comparing norms, this is onlypossible when t = 1/2, and so h(x) = −x, which contradicts that h maps no point xto its antipode −x.

Thus, we see that h is homotopic to the identity. Since the identity has degree 1,we have deg(h) = 1 by (i), a contradiction.

Proof of (d). Suppose such a v(x) exists. Then, let h(x) = v(x)/‖v(x)‖; h(x) then isa map Sn → Sn. Since v(x) is a tangent vector field, 〈h(x), x〉 = 〈v(x), x〉 /‖v(x)‖ =0, where 〈·, ·〉 is the standard inner product in Rn+1. Then, there are no pointsh(x) = ±x since 〈h(x), x〉 = 0 6= ±1. By the proofs of (b) and (c), h(x) is thenhomotopic to the identity and the antipodal map; hence, the identity is homotopicto the antipodal map. By (i), (iii), and the proofs of (b) and (c), this then impliesthat deg 1 = deg(−1)n+1, and so n is odd.

59 The Fundamental Group of Sn

Exercise 59.1. Let X be the union of two copies of S2 having a single point incommon. What is the fundamental group of X? Prove that your answer is correct.

43

Solution. Let U = X \ a, V = X \ b such that a, b are in different copies of S2;U, V are open since X is Hausdorff. Then, since S2 \ a, S2 \ b are homeomorphicto R2 by the proof in Theorem 59.3, we see that there exists a deformation retractionfrom X \ a onto S2 by taking the copy of S2 containing a and retracting it intox0 the intersection of the two copies of S2; likewise, there exists a deformationretraction from X \ b onto S2. Thus, both U, V are simply connected by Theorem59.3. U ∩V is path connected since it is homeomorphic to two copies of R2 adjoinedat x0, and so X is simply connected by Corollary 59.2.

60 Fundamental Groups of Some Surfaces

Exercise 60.2. Let X be the quotient space obtained from B2 by identifying eachpoint x of S1 with its antipode −x. Show that X is homeomorphic to the projectiveplane P 2.

Proof. Consider X as constructed from B2 identified with the closed upper hemi-sphere of S2. Let p : S2 → P 2, q : B2 → X be the quotient maps, and π : S2 → B2

the map sending x to either itself or −x if x /∈ B2; note this is a quotient map sinceU ⊆ S2 open implies π(U) open, and V ⊆ B2 open implies π−1(V ) = V ∪−V open.We then have the commutative diagram

S2 B2

P 2 X

p

π

q

f

and since q π is a quotient map by p. 141, we see f is a quotient map by Theorem22.2. But since for x ∈ X, (qπ)−1(x) = x,−x ∈ P 2, and moreover any equivalenceclass x,−x can be realized as the inverse image of this type, f is a bijection andP 2 = (q π)−1(x) | x ∈ X, and so f is a homeomorphism by Corollary 22.3(a).

Exercise 60.3. Let p : E → X be the map constructed in the proof of Lemma 60.5.Let E ′ be the subspace of E that is the union of the x-axis and the y-axis. Show thatp|E′ is not a covering map.

Proof. Consider the base point x0, the center of the figure-eight X. A neighborhoodU 3 x0 contains the union V of open intervals in A and B which intersect in exactlyx0. (p|E′)−1(V ) is then equal to the union of open intervals around integers on thex-axis and open intervals around integers on the y-axis. But none of these intervalsare homeomorphic to V , since removing an integer in an interval gives two connectedcomponents, while removing its image x0 in V gives four connected components.

44

11 The Seifert-van Kampen Theorem

67 Direct Sums of Abelian Groups

Exercise 67.2. Show that if G1 is a subgroup of G, there may be no subgroup G2 ofG such that G = G1 ⊕G2.

Proof. Let G = Z and G1 = 2Z, and suppose that such a G2 exists. Then, we haveG = 2Z⊕Z/2Z by Corollary 67.3, and so (0, 1) ∈ G1⊕G2. But since 2(0, 1) = (0, 0),G1 ⊕G2 contains an element of order 2 while G does not, a contradiction.

Exercise 67.4. The order of an element a of an abelian group G is the smallestpositive integer m such that ma = 0, if such exists; otherwise, the order of a is saidto be infinite. The order of a thus equals the order of the subgroup generated by a.

(a) Show the elements of finite order in G form a subgroup of G, called its torsionsubgroup.

(b) Show that if G is free abelian, it has no elements of finite order.(c) Show the additive group of rationals has no elements of finite order, but is not

free abelian.

Proof of (a). It suffices to show the torsion elements T (G) of G is closed under sumsand inverses. If a, b ∈ T (G) with orders m,n respectively, then mn(a+b) = n(ma)+m(nb) = 0 =⇒ a+ b ∈ T (G); likewise, m(−a) = −(ma) = 0 =⇒ −a ∈ T (G).

Proof of (b). Any 0 6= a ∈ G is of the form a =∑kαaα where aα is our basis of

G and kα ∈ Z. If ma = 0, then mkαaα = 0 for all α, which is a contradiction sinceeach aα also has infinite order.

Proof of (c). Suppose Q = 〈aα〉, aα a basis. Then for any aα, aα/2 must be of theform aα/2 =

∑kαaα for kα ∈ Z. But then, aα = 2

∑kαaα, and since aα is a basis

element, we have aα = 2kαaα. This implies kα = 1/2, which is a contradiction.

68 Free Products of Groups

Exercise 68.2. Let G = G1 ∗G2, where G1 and G2 are nontrivial groups.(a) Show G is not abelian.(b) If x ∈ G, define the length of x to be the length of the unique reduced word in

the elements of G1 and G2 that represents x. Show that if x has even length(at least 2), then x does not have finite order. Show that if x has odd length(at least 3), then x is conjugate to an element of shorter length.

45

(c) Show that the only elements of G that have finite order are the elements of G1

and G2 that have finite order, and their conjugates.

Proof of (a). For 1 6= g ∈ G1, 1 6= h ∈ G2, gh 6= hg since otherwise we would havedistinct reduced word representations of the same element of G.

Proof of (b). If x ∈ G has even length, then without loss of generality, x starts withan element in G1 and ends with one in G2, and so we cannot reduce xn to the identity,and x has infinite order. If x ∈ G has odd length, then without loss of generality,x = ghg′ for g, g′ ∈ G1, h ∈ G. Then, g−1xg = hg′g has shorter length, for g′greduces to one element in G1.

Proof of (c). Suppose x ∈ G has finite order. Then, by (b) it must have odd length2k + 1. We proceed by induction on k. For k = 0, we see the length is 1, and sox ∈ Gi for some i, and has finite order in Gi. Now suppose k > 0. Since x has oddlength, x = g−1yg for g, y ∈ G, y of shorter length. y has odd length for if not, y hasinfinite order by (b), and so x also has infinite order since xn = g−1yng, which is acontradiction. Since y is of finite order by the fact gxng−1 = yn, y is either equal toan element of Gi with finite order or conjugate to one by inductive hypothesis. If thelatter is true, x = g−1yg = g−1h−1zhg = (hg)−1zhg for h ∈ G, and z ∈ Gi havingfinite order for some i. x is therefore conjugate to a finite order element of Gi.

71 The Fundamental Group of a Wedge of Circles

Exercise 71.2. Suppose X is a space that is the union of the closed subspaces X1,. . . , Xn; assume there is a point p of X such that Xi ∩ Xj = p for i 6= j. Thenwe call X the wedge of the spaces X1, . . . , Xn, and write X = X1 ∨ · · · ∨ Xn.Show that if for each i, the point p is a deformation retract of an open set Wi ofXi, then π1(X, p) is the external free product of the groups π1(Xi, p) relative to themonomorphisms induced by inclusion.

Proof. By induction, it suffices to show when X = X1 ∨X2; moreover, it suffices toconsider when the Xi are both path-connected since if Ci are the path componentscontaining p in Xi, then π1(Ci, p) = π1(Xi, p) as on p. 332. So, let U = X1 ∪W2

and V = X2 ∪W1. Then, both U and V are path connected since they deformationretract to X1, X2, respectively, and U ∩ V = W1 ∪W2 is moreover simply connectedsince it deformation retracts to p. Thus, by Corollary 70.3, there is an isomorphismπ1(X1, p) ∗ π1(X2, p) ' π1(X, p).

Exercise 71.4. Show that if X is an infinite wedge of circles, then X does not satisfythe first countability axiom.

46

Proof. Let p ∈ X be the common point of the circles. Suppose X has a countablebasis Ui at p; we can assume without loss of generality that Ui ) Uj if i < j. Foreach Ui, we know that Vij = Ui ∩ Sj is open for any i, j by the coherence condition,and is nonempty since p ∈ Vij.

Now take V =⋃Vii; we claim it is a neighborhood of p that does not contain

any Ui. It is open by coherence since V ∩Si = Ui is open for all i, and contains p byconstruction above. Now suppose Ui ⊆ V . Then, this implies Ui ∩ Sj ⊆ V ∩ Sj = Ujfor all j, and in particular when i < j, which contradicts that Ui ) Uj if i < j.

Exercise 71.5. Let Sn be the circle of radius n in R2 whose center is at the point(n, 0). Let Y be the subspace of R2 that is the union of these circles; let p be theircommon point.

(a) Show that Y is not homeomorphic to a countably infinite wedge X of circles,nor to the space of Example 1.

(b) Show, however, that π1(Y, p) is a free group with [fn] as a system of freegenerators, where fn is a loop representing a generator of π1(Sn, p).

Proof of (a). Y is a subspace of R2 and so it is first countable by Theorem 30.2, andso Y is not homeomorphic to X by Exercise 71.4, which says X is not first countable.

Denote Z as the space of Example 71.1; we claim Z is compact while Y is not.First, denoting Dm as the closed disc of radius 1/m with center at the point (1/m, 0),we see that Z ∪ Dm is closed since it is a finite union of closed sets Z ∪ Dm =Dm ∪

⋃n<mCn. Then, Z ⊆

⋂m(Z ∪ Dm) trivially; moreover, since if x /∈ Z, then

x /∈ Dm for all m large enough, we have that Z =⋂m(Z ∪ Dm). Thus, Z is

closed. Since Z is bounded, it is then compact by Theorem 27.3. On the otherhand, Y is unbounded hence not compact by Theorem 27.3, and so Y, Z are nothomeomorphic.

Proof of (b). Let in : π1(Sn, p) → π1(Y, p) be the homomorphism induced by inclu-sion, and let their respective images be Gn; we want to show that the homomorphism

∞∗n=1

π1(Sn, p)→ π1(Y, p),J∏j=1

[fmj]`j 7→

J∏j=1

in([fmj])`j (1)

is an isomorphism.We first show that the homomorphism (1) is surjective. So, suppose f : I → Y

is a loop in Y . Letting YN = Y \⋃∞n=N+1(2n, 0), since I is compact, the image

f(I) is also compact, thus bounded by Theorem 27.3, and so f(I) ⊆ YN for some N .Letting rN : Y →

⋃Nn=1 Sn, we form the deformation retraction H : YN × I → YN ,

where H(x, 0) = idYN and H(x, 1) = rN |YN , by retracting the upper and lower

47

semicircles of Sn to p for all n > N . Then, H (f × idI) : I × I → YN is a pathhomotopy from f to rN f . Thus, [f ] = [rN f ], and Theorem 71.1 implies that[rN f ] is a product of elements of the groups Gn for n ≤ N . Therefore, [f ] is in theimage of the homomorphism (1).

We now show that the homomorphism (1) is injective. Now suppose there is somenon-identity element

w ∈∞∗n=1

π1(Sn, p),

whose image through the homomorphism (1) is the identity in π1(Y, p). Let f be aloop in X whose path-homotopy class is the image of w. Then, f is path homotopic toa constant in X, so by the argument above, it is path homotopic to a constant in someYN , and therefore path homotopic to a constant in

⋃Nn=1 Sn. But this contradicts

Theorem 71.1, which says that the map

N∗n=1

π1(Sn, p)→ π1

(N∨n=1

Sn, p

)

is injective.

73 The Fundamental Groups of the Torus and the DunceCap

Exercise 73.1. Find spaces whose fundamental groups are isomorphic to the follow-ing groups. (Here Z/n denotes the additive group of integers modulo n.)

(a) Z/n× Z/m.(b) Z/n1 × Z/n2 × · · · × Z/nk.(c) Z/n ∗ Z/m. (See Exercise 2 of §71).(d) Z/n1 ∗ Z/n2 ∗ · · · ∗ Z/nk.

Solution. We use Theorem 60.1 to say π1(X × Y, x0 × y0) = π1(X, x0) × π1(Y, y0),and use Exercise 71.2 to say π1(X ∨ Y, x0) = π1(X, x0) ∗ π1(Y, x0) for x0 = X ∩ Y .

We denote the n-fold dunce cap by Dn. Note since the Dn are path connected,we do not need to specify base points.

Through repeated applications of the above, and by using induction to get (a)⇒

48

(b), (c)⇒ (d), we get

π1(Dn ×Dm) = Z/n× Z/m =⇒ π1

(k∏i=1

Dni

)=

k∏i=1

Z/ni,

π1(Dn ∨Dm) = Z/n ∗ Z/m =⇒ π1

(k∨i=1

Dni

)=

k∗i=1

Z/ni.

12 Classification of Surfaces

74 Fundamental Groups of Surfaces

Exercise 74.3. The Klein bottle K is the space obtained form a square by meansof the labeling scheme aba−1b. Figure 74.11 indicates how K can be pictured as animmersed surface in R3.

(a) Find a presentation for the fundamental group of K.(b) Find a double covering map p : T → K, where T is the torus. Describe the

induced homomorphism of fundamental groups.

Proof of (a). π1(K) = 〈a, b | aba−1b = 1〉 by Theorem 74.2.

Proof of (b). We consider T as [0, 1]× [0, 1] with the relations (0, y) ∼ (1, y), (x, 0) ∼(x, 1), and K as [0, 1] × [0, 1] with the relations (0, y) ∼ (1, 1 − y), (x, 0) ∼ (x, 1).Then, define p : T → K by

p(x, y) =

(2x, y) if x ∈ [0, 1/2],

(2x− 1, 1− y) if x ∈ [1/2, 1].

This is continuous in each region, and agrees on the boundary since p(1/2, y) =(2 · 1/2, y) = (1, y) = (0, 1 − y) = (2 · 1/2 − 1, 1 − y) and p(1, 1 − y) = (1, 1 − y) =(0, y) = p(0, y).

Now recall that π1(T ) = 〈α, β | αβα−1β−1 = 1〉. Looking at Figures 74.4 and74.11, we see p∗(α) = a2, p∗(β) = b. Since aba−1b = 1 ⇐⇒ bab = a, we see thatp∗(α)p∗(β) = a2b = babab = ba2 = p∗(β)p∗(α).

75 Homology of Surfaces

Exercise 75.3. Let X be the quotient space obtained from an 8-sided polygonal regionP by pasting its edges together according to the labelling scheme acadbcb−1d.

49

(a) Check that all vertices of P are mapped to the same point of the quotient spaceX by the pasting map.

(b) Calculate H1(X).(c) Assuming X is homeomorphic to one of the surfaces given in Theorem 75.5

(which it is), which surface is it?

a

ac d

d

b

b−1

c

Figure 1: Labeling of edges and identification of vertices in P .

Proof of (a). We have the identification of vertices in P as shown with solid linesin Figure 1, found by identifying heads/tails of arrows with the same label. Thus,all the vertices of P are mapped to the same point of the quotient space X by thepasting map.

Solution for (b). We apply Lemma 77.1 to the labeling scheme acadbcb−1d repeatedlyto find an equivalent labeling scheme, where the brackets show our decomposition[y0]a[y1]a[y2]:

[ ]a[c]a[dbcb−1d] ∼ aac−1dbcb−1d

[aac−1]d[bcb−1]d[ ] ∼ ddaac−1bc−1b−1

[ddaa]c−1[b]c−1[b−1] ∼ c−1c−1ddaab−1b−1.

Relabeling the edges, which is allowed by p. 460, we have the labeling schemeaabbccdd. Thus, by Theorem 74.2, the fundamental group is

π1(X, x0) = 〈a, b, c, d | a2b2c2d2 = 1〉 .

Finally, the first homology group is

H1(X) ' 〈a, b, c, d | 2a+ 2b+ 2c+ 2d = 0〉 ' Z3 × (Z/2Z).

50

Solution for (c). Theorem 75.4 says that H1(Pn) = Zn−1 × (Z/2Z). Thus, assumingthat X is homeomorphic to one of the surfaces in Theorem 75.5, we see that X ishomeomorphic to P4 = P 2#P 2#P 2#P 2 by (b).

Exercise 75.4. Let X be the quotient space obtained from an 8-sided polygonal regionP by means of the labelling scheme abcdad−1cb−1. Let π : P → X be the quotientmap.

(a) Show that π does not map all the vertices of P to the same point of X.(b) Determine the space A = π(BdP ) and calculate its fundamental group.(c) Calculate π1(X, x0) and H1(X).(d) Assuming X is homeomorphic to one of the surfaces given in Theorem 75.5,

which surface is it?

a

cb d

b−1

a

cd−1

Figure 2: Labeling of edges and identification of vertices in P .

Proof of (a). We have the identification of vertices in P as shown in Figure 2, foundby identifying heads/tails of arrows with the same label, where the solid lines areone identification and the dashed lines are another. Thus, π does not map all thevertices of P to the same point of X.

a c

b

dx0 x1

Figure 3: Sketch of A = π(BdP ).

Solution for (b). Let x0 be the point identified by the solid lines in Figure 2, and x1

the point identified by the dashed lines. Then, using Figure 2, we see that a connects

51

x0 to itself, c connects x1 to itself, and b goes from x0 to x1, while d goes from x1 tox0. Thus, we have the sketch in Figure 3 for A = π(BdP ).

We now want to calculate its fundamental group. First, we see that we canhomotopically retract the segment d into the point x0, thereby making x0 coincidewith x1; the resulting deformation retract is then the wedge sum of three circles.Thus, π1(A, x0) ' π1(S1 ∨ S1 ∨ S1, x0) = Z ∗ Z ∗ Z by Theorems 58.3 and 71.1.

Solution for (c). We proceed as in Exercise 75.3(b):

[ ]a[bcd]a[d−1cb−1] ∼ aad−1c−1b−1d−1cb−1

[aad−1c−1]b−1[d−1c]b−1[ ] ∼ b−1b−1aad−1c−1c−1d

[b−1b−1aad−1]c−1[ ]c−1[d] ∼ c−1c−1b−1b−1aad−1d

c−1c−1b−1b−1aad−1d ∼ c−1c−1b−1b−1aa

c−1c−1b−1b−1aa ∼ aabbcc

where we cancel d−1d in the penultimate step and relabel in the last step as allowedon p. 460. Thus, our fundamental group is

π1(X, x0) = 〈a, b, c | a2b2c2 = 1〉 ,

which has the first homology group

H1(X) = 〈a, b, c, d | 2a+ 2b+ 2c = 0〉 ' Z2 × (Z/2Z).

Solution for (d). Since H1(X) ' Z2 × (Z/2Z), by Theorem 75.4, we have that X ishomeomorphic to P3 = P 2#P 2#P 2, assuming that X is homeomorphic to one ofthe surfaces in Theorem 75.5.

52

List of Solved Exercises

I General Topology 3

1 Set Theory and Logic 37 Countable and Uncountable Sets 3

Exercise 7.5 . . . . . . . . . . . . 3

2 Topological Spaces and Continu-ous Functions 313 Basis for a Topology . . . . . . . 3

Exercise 13.3 . . . . . . . . . . . 3Exercise 13.5 . . . . . . . . . . . 4Exercise 13.6 . . . . . . . . . . . 4Exercise 13.7 . . . . . . . . . . . 4

16 The Subspace Topology . . . . . 5Exercise 16.8 . . . . . . . . . . . 5Exercise 16.9 . . . . . . . . . . . 6

17 Closed Sets and Limit Points . . 6Exercise 17.2 . . . . . . . . . . . 6Exercise 17.3 . . . . . . . . . . . 6Exercise 17.5 . . . . . . . . . . . 7Exercise 17.13 . . . . . . . . . . . 7Exercise 17.16 . . . . . . . . . . . 7

18 Continuous Functions . . . . . . . 8Exercise 18.1 . . . . . . . . . . . 8Exercise 18.12 . . . . . . . . . . . 8

19 The Product Topology . . . . . . 9Exercise 19.6 . . . . . . . . . . . 9

20 The Metric Topology . . . . . . . 10Exercise 20.4 . . . . . . . . . . . 10Exercise 20.5 . . . . . . . . . . . 12Exercise 20.6 . . . . . . . . . . . 12Exercise 20.8 . . . . . . . . . . . 13

21 The Metric Topology (continued) 15Exercise 21.1 . . . . . . . . . . . 15Exercise 21.2 . . . . . . . . . . . 16Exercise 21.3 . . . . . . . . . . . 16

22 The Quotient Topology . . . . . . 18Exercise 22.2 . . . . . . . . . . . 18Exercise 22.4 . . . . . . . . . . . 18Exercise 22.6 . . . . . . . . . . . 19

3 Connectedness and Compactness 2023 Connected Spaces . . . . . . . . . 20

Exercise 23.8 . . . . . . . . . . . 20

Exercise 23.11 . . . . . . . . . . . 2024 Connected Subspaces of the

Real Line . . . . . . . . . . . . . 21Exercise 24.7 . . . . . . . . . . . 21Exercise 24.8 . . . . . . . . . . . 22Exercise 24.12 . . . . . . . . . . . 22

25 Components and Local Con-nectedness . . . . . . . . . . . . . 26Exercise 25.2 . . . . . . . . . . . 26

27 Compact Subspaces of the RealLine . . . . . . . . . . . . . . . . 27Exercise 27.4 . . . . . . . . . . . 27

29 Local Compactness . . . . . . . . 27Exercise 29.4 . . . . . . . . . . . 27Exercise 29.8 . . . . . . . . . . . 28

4 Countability and Separation Ax-ioms 2830 The Countability Axioms . . . . 28

Exercise 30.4 . . . . . . . . . . . 28Exercise 30.5 . . . . . . . . . . . 29Exercise 30.8 . . . . . . . . . . . 29Exercise 30.9 . . . . . . . . . . . 29Exercise 30.17 . . . . . . . . . . . 30

31 The Separation Axioms . . . . . 30Exercise 31.3 . . . . . . . . . . . 30

32 Normal Spaces . . . . . . . . . . 31Exercise 32.1 . . . . . . . . . . . 31Exercise 32.3 . . . . . . . . . . . 31Exercise 32.4 . . . . . . . . . . . 31Exercise 32.5 . . . . . . . . . . . 32

33 The Urysohn Lemma . . . . . . . 32Exercise 33.1 . . . . . . . . . . . 32

34 The Urysohn Metrization Theo-rem . . . . . . . . . . . . . . . . . 32Exercise 34.3 . . . . . . . . . . . 32Exercise 34.5 . . . . . . . . . . . 32

36 Imbeddings of Manifolds . . . . . 33Exercise 36.1 . . . . . . . . . . . 33Exercise 36.5 . . . . . . . . . . . 33

II Algebraic Topology 35

53

9 The Fundamental Group 3551 Homotopy of Paths . . . . . . . . 35

Exercise 51.1 . . . . . . . . . . . 35Exercise 51.2 . . . . . . . . . . . 35Exercise 51.3 . . . . . . . . . . . 35

52 The Fundamental Group . . . . . 36Exercise 52.4 . . . . . . . . . . . 36

53 Covering Spaces . . . . . . . . . . 37Exercise 53.3 . . . . . . . . . . . 37Exercise 53.5 . . . . . . . . . . . 37

54 The Fundamental Group of theCircle . . . . . . . . . . . . . . . 38Exercise 54.1 . . . . . . . . . . . 38Exercise 54.4 . . . . . . . . . . . 38Exercise 54.5 . . . . . . . . . . . 38

58 Deformation Retracts and Ho-motopy Type . . . . . . . . . . . 39Exercise 58.1 . . . . . . . . . . . 39Exercise 58.2 . . . . . . . . . . . 39Exercise 58.9 . . . . . . . . . . . 40Exercise 58.10 . . . . . . . . . . . 42

59 The Fundamental Group of Sn . 43Exercise 59.1 . . . . . . . . . . . 43

60 Fundamental Groups of SomeSurfaces . . . . . . . . . . . . . . 44

Exercise 60.2 . . . . . . . . . . . 44Exercise 60.3 . . . . . . . . . . . 44

11 The Seifert-van Kampen Theorem 4567 Direct Sums of Abelian Groups . 45

Exercise 67.2 . . . . . . . . . . . 45Exercise 67.4 . . . . . . . . . . . 45

68 Free Products of Groups . . . . . 45Exercise 68.2 . . . . . . . . . . . 45

71 The Fundamental Group of aWedge of Circles . . . . . . . . . 46Exercise 71.2 . . . . . . . . . . . 46Exercise 71.4 . . . . . . . . . . . 46Exercise 71.5 . . . . . . . . . . . 47

73 The Fundamental Groups of theTorus and the Dunce Cap . . . . 48Exercise 73.1 . . . . . . . . . . . 48

12 Classification of Surfaces 4974 Fundamental Groups of Surfaces 49

Exercise 74.3 . . . . . . . . . . . 4975 Homology of Surfaces . . . . . . . 49

Exercise 75.3 . . . . . . . . . . . 49Exercise 75.4 . . . . . . . . . . . 51

54